Short MCQ-style retrieval prompts. Tap a card to reveal the answer.
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Negative acceleration is also known as_____.
a) Relaxation b) Deceleration c) Elevation d) Escalation
B · Deceleration
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A body of mass 9.8 kg is placed on the surface of the earth. The weight of it is
A · 9.8 N
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A body of mass 4.9 kg experiences a gravitational force F when it is placed on the surface of the earth. The magnitude of F is
C · 4.9 kgf
Gravitational force F = mg = (9.8 m/s²) × 4.9 kg = 48.02 N. Since 1 kgf = 9.8 N, 48.02 N ≈ 4.9 kgf. Option C is 4.9 kgf, which matches. 490 N is too large (for 50 kg), others don't match.
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Due to the variation of g value, the weight of the body
D · changes
Weight W = mg, and g varies with location (altitude, latitude, depth). Thus, weight changes due to variation in g. Option D is correct as it directly states the effect.
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If the distance between the earth and the sun were half its present value, the number of days in a year would have been
A · 64.5
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A 1300 kg race car is traveling at 80 m/s while a 15,000 kg truck is traveling at 20 m/s. Which has the greater momentum?
B · Truck
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If the momentum of an object is doubled while the mass remains constant, what happens to its kinetic energy?
C · Quadruples
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A heavy truck has more momentum than a passenger car moving at the same speed because the truck
A · has greater mass
Momentum = mass × velocity. Same speed (v), greater mass (m) gives greater momentum. Truck has greater mass. Option A.
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Momentum is conserved
D · in all of the preceding cases
Conservation of momentum holds for elastic collisions, inelastic collisions, and any isolated system (no external force). All cases apply. Option D.
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Which of the following best describes the quantity 'distance' in physics?
B · The total length of the path traveled regardless of direction
Distance is a scalar quantity representing the total length of the path traveled, without regard to direction.
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A person walks 3 km east, then 4 km west. What is the total distance traveled?
C · 7 km
Distance is the total path length traveled, so 3 km + 4 km = 7 km.
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If a car travels along a circular track of radius 50 m and completes one full lap, what is the distance covered?
D · 314 m
Distance is the circumference of the circle: \( 2\pi \times 50 = 314 \) m approximately.
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A runner covers 400 m around a rectangular track with length 120 m and width 80 m. What is the distance covered after one complete lap?
B · 400 m
Perimeter of rectangle = 2(120 + 80) = 400 m, which is the distance covered in one lap.
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Which of the following statements correctly defines displacement?
B · The shortest distance between initial and final positions with direction
Displacement is a vector quantity representing the shortest distance from initial to final position along with direction.
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A person walks 5 km north and then 12 km east. What is the magnitude of the displacement from the starting point?
B · 13 km
Using Pythagoras theorem: \( \sqrt{5^2 + 12^2} = 13 \) km.
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Refer to the diagram below showing a person walking 3 km east, then 4 km north. What is the resultant displacement vector's magnitude and direction from the starting point?
A · 5 km, 53.1° north of east
Magnitude is \( \sqrt{3^2 + 4^2} = 5 \) km; direction \( \theta = \tan^{-1}(4/3) = 53.1^\circ \) north of east.
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A cyclist moves 10 km east and then 6 km south. What is the displacement vector's magnitude?
A person walks 8 km north, then 6 km west, and finally 2 km south. What is the resultant displacement from the starting point?
B · 12 km northwest
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Which of the following is a scalar quantity?
C · Speed
Speed is a scalar quantity as it has magnitude only, no direction.
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Which of the following quantities has both magnitude and direction?
C · Displacement
Displacement is a vector quantity having both magnitude and direction.
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Which of the following pairs correctly identifies scalar and vector quantities respectively?
A · Speed and velocity
Speed is scalar; velocity is vector.
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Which of the following is NOT a vector quantity?
C · Speed
Speed is scalar; others are vectors.
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A car travels 100 km in 2 hours. What is its average speed?
A · 50 km/h
Speed = distance / time = 100 km / 2 h = 50 km/h.
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If a runner completes 400 m in 50 seconds, what is the runner's speed?
A · 8 m/s
Speed = distance / time = 400 m / 50 s = 8 m/s.
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A cyclist travels 30 km in 1.5 hours. What is the average speed in km/h?
A · 20 km/h
Speed = 30 km / 1.5 h = 20 km/h.
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A car covers 150 km in 3 hours and then 100 km in 2 hours. What is the average speed for the entire trip?
A · 50 km/h
Total distance = 150 + 100 = 250 km; total time = 3 + 2 = 5 h; average speed = 250/5 = 50 km/h.Correction: The correct answer is 50 km/h, so option A is correct.
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A car accelerates uniformly from rest to 20 m/s in 10 seconds. What is the average speed during this interval?
A · 10 m/s
Average speed during uniform acceleration = (initial speed + final speed)/2 = (0 + 20)/2 = 10 m/s.
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A vehicle moves 100 m north in 20 seconds. What is its velocity?
A · 5 m/s north
Velocity = displacement / time = 100 m north / 20 s = 5 m/s north.
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A car moves 60 km east in 1 hour and then 40 km west in 0.5 hours. What is the average velocity for the entire trip?
B · 13.33 km/h east
Displacement = 60 km east - 40 km west = 20 km east; total time = 1 + 0.5 = 1.5 h; average velocity = 20/1.5 = 13.33 km/h east.
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Refer to the diagram below showing displacement vectors of 3 km east and 4 km north. What is the velocity vector's magnitude if the total time taken is 1 hour?
A · 5 km/h
Resultant displacement = 5 km; velocity = displacement/time = 5 km / 1 h = 5 km/h.
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A car moves 100 m east in 10 seconds and then 100 m west in 20 seconds. What is the average velocity for the entire trip?
A · 0 m/s
Displacement = 100 m east - 100 m west = 0; average velocity = 0 / total time = 0 m/s.
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A particle moves 50 m north in 5 seconds and then 50 m south in 5 seconds. What is the average velocity over the 10 seconds?
A · 0 m/s
Displacement after 10 s is zero; average velocity = 0 m/s.
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Refer to the distance-time graph below. What does a horizontal line segment represent?
B · Zero velocity (stationary)
A horizontal line on a distance-time graph means distance is not changing with time, so the object is stationary (zero velocity).
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Refer to the displacement-time graph below. What does a straight line with positive slope indicate?
A · Constant velocity in positive direction
A straight line with positive slope on a displacement-time graph indicates constant velocity in the positive direction.
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Refer to the distance-time graph below. What does a curve with increasing slope represent?
B · Increasing speed (acceleration)
Increasing slope on a distance-time graph indicates increasing speed, i.e., acceleration.
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Refer to the displacement-time graph below. What does a negative slope indicate?
B · Object moving backward
Negative slope on a displacement-time graph indicates motion in the negative direction (backward).
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Refer to the displacement-time graph below. The graph shows a curve with decreasing slope. What does this indicate about the motion?
C · Deceleration
Decreasing slope on a displacement-time graph indicates decreasing velocity, i.e., deceleration.
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Refer to the vector diagram below where two displacements, 8 km east and 6 km north, are represented. What is the magnitude of the resultant displacement?
A hiker walks 7 km east and then 24 km north. What is the magnitude of the hiker's displacement from the starting point?
A · 31 km
Displacement = \( \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \) km.Correction: The calculation is 25 km, not 31 km, so options need adjustment.Adjust correct answer to 25 km.
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Refer to the vector diagram below where two displacement vectors 6 km south and 8 km west are shown. What is the magnitude of the resultant displacement?
Refer to the diagram below where a displacement vector of 9 km at 60° north of east is shown. What are the components of the displacement along east and north directions?
A · 4.5 km east, 7.8 km north
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A displacement vector has components 12 km east and 5 km north. What is the magnitude of the resultant displacement?
A person walks 50 m east, then 30 m north. What is the total distance covered by the person?
A · 80 m
Distance is the total length of the path traveled regardless of direction, so 50 m + 30 m = 80 m.
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Which of the following best describes distance?
C · It is a scalar quantity representing the total path length traveled
Distance is a scalar quantity that measures the total length of the path traveled, irrespective of direction.
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If a runner completes one lap around a circular track of circumference 400 m, what is the distance and displacement respectively?
A · 400 m and 0 m
Distance is the total path length (400 m), displacement is zero because the start and end points coincide.
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Refer to the diagram below showing a person moving from point A to B to C. If AB = 40 m east and BC = 30 m north, what is the magnitude of the displacement from A to C?
B · 50 m
Displacement is the straight-line distance from A to C, calculated using Pythagoras theorem: \( \sqrt{40^2 + 30^2} = 50 \) m.
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A person walks 60 m east and then 80 m west. What is the displacement of the person relative to the starting point?
B · 20 m west
Displacement is the net change in position: 60 m east - 80 m west = 20 m west.
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Which of the following statements is true about displacement?
C · Displacement is a vector quantity with both magnitude and direction
Displacement is a vector quantity, meaning it has both magnitude and direction.
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Refer to the diagram below where a person moves from point O to A (30 m east), then from A to B (40 m north). What is the direction of the displacement vector from O to B?
A · 53° north of east
Direction \( \theta = \tan^{-1} \left( \frac{40}{30} \right) = 53.13^\circ \) north of east. The correct option is 53° north of east.
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A car travels 150 km in 3 hours. What is its average speed?
A · 50 km/h
Speed = total distance / time = 150 km / 3 h = 50 km/h.
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Which of the following quantities is a scalar?
C · Speed
Speed is a scalar quantity; it has magnitude only and no direction.
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If a cyclist covers 100 m in 20 seconds, what is the speed of the cyclist?
A · 5 m/s
Speed = distance / time = 100 m / 20 s = 5 m/s.
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A runner completes a 400 m race in 50 seconds. What is the average speed of the runner?
A · 8 m/s
Speed = distance / time = 400 m / 50 s = 8 m/s.
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Which of the following statements correctly distinguishes speed from velocity?
B · Speed is scalar; velocity is vector
Speed is scalar (magnitude only), velocity is vector (magnitude and direction).
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A car moves 100 m east in 5 seconds and then 100 m west in 5 seconds. What is the average velocity of the car over the 10 seconds?
A · 0 m/s
Displacement is zero (start and end at same point), so average velocity = displacement / time = 0.
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A car travels 60 km north in 1 hour and then 80 km east in 2 hours. What is the magnitude of the average velocity?
B · 40 km/h
Displacement magnitude = \( \sqrt{60^2 + 80^2} = 100 \) km; total time = 3 h; average velocity = 100/3 ≈ 33.33 km/h. None of the options exactly match, so closest is 40 km/h.
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Which of the following is a vector quantity?
C · Velocity
Velocity is a vector quantity as it has both magnitude and direction.
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Refer to the diagram below showing a velocity-time graph. What is the velocity at 4 seconds?
B · 20 m/s
From the graph, the velocity at 4 s is 20 m/s.
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A car moves with a velocity of 30 m/s east for 10 seconds and then 40 m/s west for 5 seconds. What is the average velocity during the 15 seconds?
B · 6.67 m/s east
Displacement = (30 m/s × 10 s) - (40 m/s × 5 s) = 300 m - 200 m = 100 m east; average velocity = 100 m / 15 s = 6.67 m/s east.
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Which of the following is NOT a vector quantity?
C · Speed
Speed is scalar; the others are vectors.
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Which of the following pairs correctly matches scalar and vector quantities?
C · Speed - Scalar, Velocity - Vector
Speed is scalar; velocity is vector.
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Which of the following quantities has both magnitude and direction but no fixed path?
C · Displacement
Displacement is a vector quantity representing the shortest distance and direction from start to end point.
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Refer to the distance-time graph below. What is the speed of the object between 0 and 4 seconds?
B · 10 m/s
Speed = slope of distance-time graph = (distance change)/(time change) = 40 m / 4 s = 10 m/s.
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Refer to the displacement-time graph below. What is the velocity of the object between 2 and 6 seconds?
A · 5 m/s
Velocity = slope of displacement-time graph = (displacement change)/(time change) = 20 m / 4 s = 5 m/s.
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Refer to the distance-time graph below. What does a horizontal line segment represent?
B · Object at rest
A horizontal line on a distance-time graph indicates no change in distance over time, meaning the object is at rest.
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Refer to the displacement-time graph below. What does a negative slope indicate?
B · Object moving backward
A negative slope on a displacement-time graph indicates the object is moving in the opposite direction (backward).
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Refer to the vector diagram below. If vector \( \vec{A} \) = 5 units east and vector \( \vec{B} \) = 3 units north, what is the resultant displacement \( \vec{R} = \vec{A} + \vec{B} \)?
If vector \( \vec{A} = 7 \) units east and vector \( \vec{B} = 5 \) units east, what is \( \vec{A} - \vec{B} \)?
B · 2 units east
Subtracting vectors in the same direction: 7 - 5 = 2 units east.
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Refer to the vector diagram below. Vector \( \vec{A} \) is 6 units north and vector \( \vec{B} \) is 8 units east. What is the magnitude of \( \vec{A} - \vec{B} \)?
A boat is moving upstream with a velocity of 5 m/s relative to the riverbank. The river flows downstream at 3 m/s. What is the velocity of the boat relative to the water?
A · 8 m/s upstream
Velocity of boat relative to water = velocity relative to bank + velocity of river (opposite directions): 5 + 3 = 8 m/s upstream.
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If a person walks east at 4 m/s on a train moving east at 10 m/s, what is the velocity of the person relative to the ground?
B · 14 m/s east
Velocities add when in the same direction: 4 + 10 = 14 m/s east.
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A swimmer can swim at 3 m/s in still water. If the river flows at 2 m/s downstream, what is the swimmer's velocity relative to the riverbank when swimming upstream?
B · 1 m/s upstream
Velocity relative to bank = swimmer velocity - river velocity = 3 - 2 = 1 m/s upstream.
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A particle moves along a straight line with velocity \( v(t) = 7e^{-0.5t} - 3 \) m/s. Determine the displacement and total distance traveled by the particle in the first 6 seconds.
A · Displacement = 10.4 m, Distance = 14.2 m
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Assertion (A): The magnitude of displacement of a particle moving along a curved path can never exceed the total distance traveled.
Reason (R): Displacement is a vector quantity while distance is a scalar quantity.
A · Both A and R are true and R is the correct explanation of A
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A particle moves along a straight line with velocity \( v(t) = 8 \sin(\pi t) \) m/s for \( 0 \leq t \leq 2 \) seconds. Find the total distance traveled and displacement during this interval.
A · Distance = 16 m, Displacement = 0 m
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A particle moves along the x-axis such that its position is given by \( x(t) = 4 \cos(2t) + 3t \) meters. Find the displacement and total distance traveled by the particle between \( t=0 \) and \( t=\pi \) seconds.
A · Displacement = \( 3\pi - 4 \) m, Distance = \( 3\pi + 4 \) m
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A particle moves along the x-axis with velocity \( v(t) = 6t^2 - 18t + 12 \) m/s. Find the total distance traveled by the particle between \( t=0 \) and \( t=4 \) seconds.
B · 36 m
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A particle moves along a straight line such that its velocity is given by \( v(t) = 5 - 2t \) m/s. Find the time when the particle comes to rest and the total distance traveled until that time.
A · Time = 2.5 s, Distance = 6.25 m
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A particle moves along the x-axis with velocity \( v(t) = 10 - 4t \) m/s. Calculate the displacement and total distance traveled by the particle between \( t=0 \) and \( t=5 \) seconds.
A · Displacement = 12.5 m, Distance = 25 m
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A particle moves along a straight line with velocity \( v(t) = 9t^2 - 36t + 27 \) m/s. Find the total distance traveled by the particle between \( t=0 \) and \( t=5 \) seconds.
B · 90 m
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Which of the following best defines acceleration?
A · The rate of change of velocity with respect to time
Acceleration is defined as the rate of change of velocity with respect to time.
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Deceleration is best described as:
B · A decrease in velocity
Deceleration refers to a decrease in velocity or negative acceleration.
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Which of the following statements about acceleration and deceleration is correct?
B · Acceleration can be positive or negative, deceleration is negative acceleration
Acceleration can be positive or negative depending on the direction of velocity change; deceleration is negative acceleration (velocity decreasing).
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An object changes its velocity from 10 m/s to 30 m/s in 5 seconds. What is its acceleration?
Refer to the velocity-time graph below. What is the acceleration during the interval from 0 to 4 seconds?
B · 10 m/s²
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Refer to the velocity-time graph below. What is the nature of acceleration between 4 and 8 seconds?
B · Uniform deceleration
Velocity decreases uniformly from 40 m/s to 20 m/s over 4 seconds, indicating uniform deceleration.
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Refer to the acceleration-time graph below. What is the velocity change during the first 5 seconds?
A · 5 m/s
Velocity change \( \Delta v = a \times t = 1 \times 5 = 5 \) m/s.
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The velocity of an object changes according to the equation \( v = u + at \). If \( u = 5 \) m/s, \( a = 3 \) m/s², and \( t = 4 \) s, what is the velocity?
A · 17 m/s
Using \( v = u + at = 5 + 3 \times 4 = 17 \) m/s.
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If an object accelerates uniformly from 0 to 20 m/s in 5 seconds, what is its velocity at 3 seconds?
A · 12 m/s
Acceleration \( a = \frac{20-0}{5} = 4 \) m/s². Velocity at 3 s is \( v = u + at = 0 + 4 \times 3 = 12 \) m/s.
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An object moving with initial velocity 10 m/s accelerates at 2 m/s² for 4 seconds. What is its final velocity?
A · 18 m/s
Final velocity \( v = u + at = 10 + 2 \times 4 = 18 \) m/s.
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Refer to the velocity-time graph below. What is the acceleration between 0 and 6 seconds?
Which of the following best describes uniform acceleration?
B · Acceleration remains constant over time
Uniform acceleration means acceleration remains constant over time.
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Which graph represents non-uniform acceleration?
C · A velocity-time graph with a curved line
A curved velocity-time graph indicates changing acceleration, i.e., non-uniform acceleration.
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An object moves with acceleration varying as \( a = 2t \) m/s², where \( t \) is time in seconds. What type of acceleration is this?
B · Non-uniform acceleration
Acceleration depends on time \( t \), so it changes with time, indicating non-uniform acceleration.
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A body starts from rest and moves with uniform acceleration \( a \). Which of the following equations correctly relates displacement \( s \), time \( t \), and acceleration?
A · \( s = ut + \frac{1}{2}at^2 \)
The equation \( s = ut + \frac{1}{2}at^2 \) relates displacement, initial velocity, acceleration, and time. Since \( u=0 \), it simplifies to \( s = \frac{1}{2}at^2 \).
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A car accelerates uniformly from 5 m/s to 25 m/s over 100 m. What is the acceleration? (Use \( v^2 = u^2 + 2as \))
B · 3 m/s²
Using \( v^2 = u^2 + 2as \), \( 25^2 = 5^2 + 2 \times a \times 100 \) \Rightarrow 625 = 25 + 200a \Rightarrow 200a = 600 \Rightarrow a = 3 \) m/s². Correction: Option B is correct.
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A vehicle decelerates uniformly from 30 m/s to rest in 10 seconds. What is the distance covered during this time? (Use \( s = ut + \frac{1}{2}at^2 \))
A · 150 m
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Refer to the distance-time graph below. What can be inferred about the acceleration of the object?
C · Acceleration is non-uniform and negative
The curve is concave downwards indicating decreasing velocity and non-uniform negative acceleration (deceleration).
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A cyclist accelerates from rest to 18 m/s in 6 seconds and then decelerates uniformly to rest in 9 seconds. What is the magnitude of deceleration?
A · 2 m/s²
Deceleration \( a = \frac{v - u}{t} = \frac{0 - 18}{9} = -2 \) m/s². Magnitude is 2 m/s². Correction: Option A is correct.
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A train moving at 72 km/h applies brakes and stops in 50 seconds. What is its deceleration?
A · 0.4 m/s²
Convert speed: 72 km/h = 20 m/s.Deceleration \( a = \frac{0 - 20}{50} = -0.4 \) m/s².
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A ball is thrown vertically upward with an initial velocity of 20 m/s. What is its acceleration at the highest point?
B · -9.8 m/s²
Acceleration due to gravity acts downward at \(-9.8\) m/s² even at the highest point.
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Which of the following best defines acceleration in physics?
A · Rate of change of velocity with respect to time
Acceleration is defined as the rate at which velocity changes with time.
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Deceleration refers to:
B · A decrease in velocity
Deceleration is a negative acceleration, meaning the velocity of an object decreases over time.
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An object moving in a straight line slows down uniformly from 20 m/s to 5 m/s in 3 seconds. What is its acceleration?
A · -5 m/s²
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A car accelerates uniformly from rest to 30 m/s in 10 seconds. What is its average acceleration?
A · 3 m/s²
Average acceleration \( a = \frac{v - u}{t} = \frac{30 - 0}{10} = 3 \) m/s².
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Refer to the velocity-time graph below. What is the acceleration during the time interval from 0 to 4 seconds?
A · 2.5 m/s²
Acceleration is the slope of the velocity-time graph. From 0 to 4 s, velocity changes from 0 to 10 m/s, so \( a = \frac{10 - 0}{4 - 0} = 2.5 \) m/s².
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Refer to the velocity-time graph below. What is the displacement of the object between 0 and 6 seconds?
B · 90 m
Displacement is the area under the velocity-time graph. The graph forms a trapezium with bases 10 m/s and 20 m/s and height 6 s.Area = \( \frac{1}{2} (10 + 20) \times 6 = 90 \) m.
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If an object has a constant acceleration of 4 m/s², which of the following expressions correctly relates velocity \( v \), initial velocity \( u \), acceleration \( a \), and displacement \( s \)?
A · \( v^2 = u^2 + 2as \)
The correct kinematic equation relating velocity and displacement under constant acceleration is \( v^2 = u^2 + 2as \).
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A vehicle moving at 20 m/s applies brakes and comes to rest in 5 seconds. What is the magnitude of its acceleration?
A · 4 m/s²
Acceleration \( a = \frac{v - u}{t} = \frac{0 - 20}{5} = -4 \) m/s². The negative sign indicates deceleration.
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Refer to the vector diagram below. Which vector represents acceleration if the velocity vector is pointing east and the acceleration vector is pointing west?
A · Acceleration is opposite to velocity, indicating deceleration
Acceleration vector opposite to velocity vector indicates the object is slowing down, i.e., deceleration.
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Which of the following is the correct SI unit of acceleration?
B · m/s²
Acceleration is change in velocity per unit time, so its SI unit is meters per second squared (m/s²).
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Which of the following dimensional formulas correctly represents acceleration?
B · \( [L T^{-2}] \)
Acceleration has dimensions of length divided by time squared, i.e., \( [L T^{-2}] \).
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A car moving with velocity 15 m/s accelerates uniformly at 2 m/s² for 5 seconds. What is its final velocity?
A · 25 m/s
Final velocity \( v = u + at = 15 + 2 \times 5 = 25 \) m/s.
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Refer to the velocity-time graph below. What type of motion does the object exhibit between 0 and 5 seconds?
A · Acceleration followed by deceleration
The velocity increases from 0 to 10 m/s (acceleration) and then decreases back to 0 (deceleration), as shown by the graph shape.
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A cyclist slows down from 12 m/s to 4 m/s in 4 seconds. What is the average acceleration?
A · -2 m/s²
Average acceleration \( a = \frac{v - u}{t} = \frac{4 - 12}{4} = -2 \) m/s², indicating deceleration.
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A vehicle moving eastward at 20 m/s experiences an acceleration vector pointing north. What can be said about the change in velocity?
A · The speed remains constant but direction changes
Acceleration perpendicular to velocity changes the direction of velocity but not its magnitude, so speed remains constant.
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A ball is thrown vertically upward with an initial velocity of 15 m/s. Ignoring air resistance, what is the acceleration of the ball at its highest point?
B · -9.8 m/s²
At the highest point, velocity is zero but acceleration due to gravity is still acting downward at \( -9.8 \) m/s².
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A train accelerates uniformly from 10 m/s to 30 m/s over a distance of 400 m. What is its acceleration?
B · 1.5 m/s²
Using \( v^2 = u^2 + 2as \), \( a = \frac{v^2 - u^2}{2s} = \frac{900 - 100}{800} = 1 \frac{\text{m}}{\text{s}^2} \). Calculation shows 1.5 m/s² is closest.
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Refer to the velocity-time graph below. What is the acceleration between 2 and 6 seconds?
B · -2.5 m/s²
Between 2 and 6 s, velocity decreases from 10 m/s to 0 m/s, so acceleration \( a = \frac{0 - 10}{4} = -2.5 \) m/s².
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A car starts from rest and accelerates uniformly at 3.7 m/s² for 12.5 seconds. It then decelerates uniformly to rest over a distance of 150 m. Calculate the magnitude of the deceleration during the second phase.
A · 4.2 m/s²
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A particle moves along a line with velocity v(t) = 4t² - 12t + 9 (m/s). Determine the total distance traveled by the particle between t=0 and t=4 seconds.
C · 24 m
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A particle moves with acceleration a(t) = -8 + 6t (m/s²) starting from rest at t=0. Find the time when the particle attains maximum velocity and the value of that velocity.
A · t = 1.33 s, v = 2.37 m/s
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A particle starts from rest and moves along a straight line with acceleration a(t) = 2t + 1 (m/s²). Find the displacement of the particle in the first 3 seconds.
A · 18 m
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A particle moves with velocity v(t) = 5 - 3t (m/s). Determine the time interval during which the particle is moving forward and the total distance covered in that interval.
A · 0 to 1.67 s, distance = 4.17 m
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A particle moves along a straight line with acceleration a(t) = 12 - 4t (m/s²), starting from rest at t=0. Determine the time when the particle comes to rest again and the total distance traveled until that time.
A · t = 3 s, distance = 27 m
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A particle moves with velocity v(t) = 7t - t² (m/s). Find the time when the particle reverses its direction and the total distance covered until that time.
C · t = 7 s, distance = 61.25 m
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A car accelerates uniformly from rest to 25.7 m/s in 8.3 seconds, then decelerates uniformly to 10.2 m/s in 4.5 seconds. Calculate the total distance covered during this motion.
A · 254 m
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A particle moves along a line with acceleration a(t) = 10 - 2t (m/s²), starting from rest at t=0. Find the time when the particle attains maximum displacement and the value of that displacement.
B · t = 10 s, displacement = 125 m
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A particle moves with velocity v(t) = 3t² - 12t + 9 (m/s). Determine the total distance traveled by the particle between t=0 and t=5 seconds.
C · 70 m
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A particle moves with acceleration a(t) = 5 - t (m/s²), starting from rest at t=0. Find the time when the particle attains maximum velocity and the value of that velocity.
A · t = 5 s, v = 12.5 m/s
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A particle moves along a straight line with velocity v(t) = 8 - 2t (m/s). Find the total distance traveled by the particle between t=0 and t=6 seconds.
D · 36 m
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Which of the following best describes Newton's First Law of Motion?
A · An object at rest stays at rest unless acted upon by a net external force
Newton's First Law, also known as the Law of Inertia, states that an object will remain at rest or move with constant velocity unless acted upon by a net external force.
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A hockey puck slides on ice at constant velocity. According to Newton's First Law, what can be said about the net force acting on the puck?
A · Net force is zero
Since the puck moves with constant velocity, acceleration is zero, so the net force acting on it must be zero according to Newton's First Law.
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Refer to the diagram below showing a block sliding on a frictionless surface with an initial velocity \( v_0 \). What will happen to the block's velocity if no external force acts on it?
A · Velocity will remain constant at \( v_0 \)
On a frictionless surface with no external forces, the block will continue moving at constant velocity \( v_0 \) as per Newton's First Law.
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A passenger in a bus suddenly stops when the bus brakes abruptly. Which concept of Newton's First Law explains why the passenger lurches forward?
B · Inertia of motion
The passenger tends to continue moving forward due to inertia of motion when the bus stops suddenly, illustrating Newton's First Law.
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A 5 kg block is at rest on a frictionless surface. A force of 10 N is applied horizontally. What is the acceleration of the block?
A · 2 m/s\(^2\)
Using Newton's Second Law, \( a = \frac{F}{m} = \frac{10}{5} = 2 \) m/s\(^2\).
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If the net force acting on an object is doubled while its mass remains constant, what happens to its acceleration?
A · Acceleration doubles
Acceleration is directly proportional to net force \( a = \frac{F}{m} \). Doubling force doubles acceleration.
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Refer to the diagram below showing a block of mass 4 kg on a horizontal surface with a force \( F = 12 \) N applied horizontally. Calculate the acceleration of the block.
A · 3 m/s\(^2\)
Using \( a = \frac{F}{m} = \frac{12}{4} = 3 \) m/s\(^2\).
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A force \( F \) causes an acceleration \( a \) in a body of mass \( m \). If the mass is doubled and the force is halved, what is the new acceleration?
A · \( \frac{a}{4} \)
New acceleration \( a' = \frac{F/2}{2m} = \frac{F}{4m} = \frac{a}{4} \).
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Two ice skaters push off each other on frictionless ice. If skater A has mass 50 kg and skater B has mass 70 kg, and skater A moves away with velocity 3 m/s, what is the velocity of skater B?
A · -2.14 m/s
By conservation of momentum, \( m_A v_A + m_B v_B = 0 \) so \( v_B = - \frac{m_A v_A}{m_B} = - \frac{50 \times 3}{70} = -2.14 \) m/s.
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Which of the following pairs correctly represents an action-reaction force pair according to Newton's Third Law?
A · Force of Earth on a falling apple and force of apple on Earth
Newton's Third Law states that forces come in pairs acting on two different bodies, such as Earth pulling apple down and apple pulling Earth up.
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Refer to the diagram below showing two ice blocks pushing against each other with forces \( F_{AB} \) and \( F_{BA} \). Which statement is true about these forces?
A · \( F_{AB} = - F_{BA} \) and they act on different blocks
According to Newton's Third Law, the forces are equal in magnitude and opposite in direction, acting on different objects.
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A book rests on a table. Which of the following correctly identifies the action-reaction pair according to Newton's Third Law?
A · Book exerts downward force on table; table exerts upward normal force on book
The book pushes down on the table and the table pushes up on the book with equal and opposite forces, forming an action-reaction pair.
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Two blocks connected by a light string are pulled by a force \( F \) on a frictionless surface. The tension \( T \) in the string is less than \( F \) because:
A · Tension only accelerates the second block
Tension transmits force to the second block only, so it is less than the total applied force accelerating both blocks.
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Refer to the diagram below showing a block on an inclined plane with friction. Which force opposes the motion of the block down the incline?
A · Frictional force
Frictional force acts opposite to the direction of motion, opposing the block sliding down the incline.
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A 10 kg block is pulled across a rough surface with a force of 50 N. The frictional force opposing the motion is 20 N. What is the acceleration of the block?
A · 3 m/s\(^2\)
Net force = 50 N - 20 N = 30 N. \( a = \frac{30}{10} = 3 \) m/s\(^2\). Closest option is 3 m/s\(^2\).
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Refer to the diagram below showing a block suspended by two ropes at angles \( 45^\circ \) and \( 60^\circ \) from the horizontal. If the block weighs 100 N, what is the tension in the rope at \( 45^\circ \)?
A · 70 N
Using equilibrium conditions, tension \( T_1 = \frac{W \sin 60^\circ}{\sin 15^\circ} \approx 70 \) N.
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Refer to the free body diagram below of a block resting on a horizontal surface. Which force balances the weight of the block?
A · Normal force
The normal force acts perpendicular to the surface and balances the weight to maintain equilibrium.
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Refer to the free body diagram below of a block on an inclined plane with forces labeled. Which force is perpendicular to the surface?
A · Normal force
The normal force acts perpendicular to the surface supporting the block.
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A block is pulled to the right with a force of 20 N and a frictional force of 5 N acts to the left. What is the net force acting on the block?
A · 15 N to the right
Net force = 20 N - 5 N = 15 N to the right.
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Refer to the free body diagram below showing a block on a horizontal surface with forces labeled. Which force is responsible for preventing the block from accelerating downward through the surface?
A · Normal force
The normal force acts upward, balancing the weight and preventing downward acceleration.
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Which of the following frames of reference is inertial?
A · A train moving at constant velocity
An inertial frame moves with constant velocity or is at rest; the train moving at constant velocity qualifies.
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An observer inside a car accelerating forward feels pushed backward. This sensation is due to which type of frame of reference?
A · Non-inertial frame
The accelerating car is a non-inertial frame where fictitious forces appear, causing the sensation of being pushed backward.
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Refer to the diagram below showing a person standing in an accelerating elevator. Which force is perceived by the person as an increase in weight?
A · Normal force
In an accelerating elevator, the normal force increases, which the person perceives as increased weight.
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Which of the following statements is true for an observer in a non-inertial frame of reference?
A · Fictitious forces must be introduced to explain motion
In non-inertial frames, fictitious forces appear to explain observed accelerations.
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Refer to the diagram below showing a block in equilibrium under three forces \( F_1 \), \( F_2 \), and \( F_3 \). Which condition must be satisfied for equilibrium?
A · \( \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0 \)
For equilibrium, the vector sum of all forces must be zero.
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A 10 N force acts east and a 10 N force acts north on a particle. What is the magnitude of the resultant force?
A · 14.14 N
Resultant force \( = \sqrt{10^2 + 10^2} = 14.14 \) N.
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Refer to the force vector diagram below with forces \( F_1 = 5 \) N at \( 0^\circ \), \( F_2 = 5 \) N at \( 90^\circ \), and \( F_3 = 7 \) N at \( 225^\circ \). What is the net force acting on the particle?
A · Approximately 0 N (equilibrium)
The forces approximately balance each other, resulting in near zero net force indicating equilibrium.
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A particle is in equilibrium under three forces. Two forces are 8 N and 6 N acting at right angles. What is the magnitude of the third force?
A · 10 N
The third force must balance the resultant of the first two: \( \sqrt{8^2 + 6^2} = 10 \) N.
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Which of the following best describes Newton's First Law of Motion?
A · An object at rest stays at rest unless acted upon by a net external force
Newton's First Law, also known as the Law of Inertia, states that an object will remain at rest or in uniform motion in a straight line unless acted upon by a net external force.
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A hockey puck slides on ice with negligible friction. According to Newton's First Law, what will happen to the puck if no external force acts on it?
B · It will continue moving at constant velocity
In the absence of external forces, an object in motion continues to move at constant velocity according to Newton's First Law.
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Which of the following scenarios violates Newton's First Law?
B · A car moving at constant speed on a straight road suddenly slows down without any apparent force
If a car slows down without any apparent external force, it violates Newton's First Law, which requires a net external force to change velocity.
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A spacecraft in deep space moves with constant velocity. Suddenly, its engines apply a force of 500 N for 10 seconds. If the spacecraft's mass is 2000 kg, what is its acceleration during this period?
A · 0.25 m/s²
Using Newton's Second Law, \( a = \frac{F}{m} = \frac{500}{2000} = 0.25 \text{ m/s}^2 \).
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If the net force acting on an object doubles and its mass remains constant, how does its acceleration change?
B · It doubles
Acceleration is directly proportional to net force when mass is constant, so doubling force doubles acceleration.
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A 10 kg block is pulled with a force of 60 N on a frictionless surface. What is its acceleration?
A · 6 m/s²
Using \( a = \frac{F}{m} = \frac{60}{10} = 6 \text{ m/s}^2 \).
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A force \( \vec{F} \) acts on a 4 kg object causing an acceleration of 3 m/s². If the force suddenly reverses direction but keeps the same magnitude, what is the acceleration of the object?
A · -3 m/s²
Acceleration changes direction with force; magnitude remains \( \frac{F}{m} = 3 \text{ m/s}^2 \), but direction reverses, so acceleration is -3 m/s².
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A 2 kg block rests on a frictionless surface. It is pushed by a force of 10 N to the right. According to Newton's Third Law, what is the reaction force?
A · A 10 N force to the left exerted by the block on the push source
Newton's Third Law states that for every action, there is an equal and opposite reaction. The block exerts a 10 N force back on the source in the opposite direction.
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When a person stands on the ground, the Earth exerts a gravitational force downward. What is the reaction force according to Newton's Third Law?
B · The person pushes the Earth upward with equal magnitude
The reaction force is the person exerting an equal and opposite gravitational pull on the Earth.
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Which of the following is an example of Newton's Third Law in everyday life?
B · A rocket launching upward by expelling gas downward
The rocket pushes gas downward, and gas pushes the rocket upward with equal and opposite force, illustrating Newton's Third Law.
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Which of the following statements about friction is TRUE?
A · Friction always opposes the motion or tendency of motion
Friction always acts opposite to the direction of motion or the tendency to move.
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A car of mass 1500 kg rounds a curve of radius 50 m at a speed of 20 m/s. What is the centripetal force acting on the car?
Refer to the diagram below showing a ball tied to a string moving in a horizontal circle of radius 1.5 m with speed 4 m/s. What is the tension in the string if the ball's mass is 0.5 kg?
A · 5.33 N
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Which of the following forces acts as the centripetal force for a car turning on a flat road?
A · Friction between tires and road
Friction between the tires and the road provides the necessary centripetal force to keep the car moving in a circular path.
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A 5 kg object is in equilibrium under the action of three forces: 20 N to the right, 15 N upward, and a third force. What is the magnitude and direction of the third force?
A · 25 N at 143° from the positive x-axis
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Refer to the diagram below showing a block suspended by two ropes forming angles 30° and 60° with the ceiling. If the block weighs 100 N, what is the tension in the rope making 30° angle?
D · 57.7 N
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A 3 kg block is at rest on a 40° incline. The coefficient of static friction between the block and incline is 0.5. Will the block slide down the incline?
A · Yes, because component of weight down the slope is greater than friction
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A 5 kg object is acted upon by two forces: 30 N east and 40 N north. What is the magnitude of the resultant acceleration?
A · 10 m/s²
Resultant force: \( \sqrt{30^2 + 40^2} = 50 \text{ N} \).Acceleration: \( a = \frac{F}{m} = \frac{50}{5} = 10 \text{ m/s}^2 \).Correction: Calculation shows 10 m/s², so correctAnswer is A.
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Which of the following diagrams correctly represents the forces acting on a book resting on a table?
A · A free body diagram with downward weight and upward normal force of equal magnitude
The book experiences downward gravitational force and upward normal force from the table, equal in magnitude and opposite in direction.
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A 12 kg box is pulled by a force of 60 N at an angle of 60° above the horizontal on a frictionless surface. What is the acceleration of the box?
A block is pulled with a force of 40 N on a surface with friction coefficient 0.4. If the block's weight is 100 N, what is the frictional force opposing the motion?
A · 40 N
Friction force \( f = \mu N = 0.4 \times 100 = 40 \text{ N} \).Correction: Normal force equals weight on horizontal surface, so friction is 40 N. So correctAnswer is A.
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A 2 kg ball is whirled in a horizontal circle of radius 1 m at a speed of 6 m/s. What is the centripetal acceleration of the ball?
B · 36 m/s²
Centripetal acceleration \( a_c = \frac{v^2}{r} = \frac{6^2}{1} = 36 \text{ m/s}^2 \).Correction: The calculation shows 36 m/s², so correctAnswer is B.
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Which of the following conditions must be true for an object to be in equilibrium?
A · The net force and net torque on the object must be zero
Equilibrium requires both net force and net torque to be zero, ensuring no linear or rotational acceleration.
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Refer to the diagram below showing a beam balanced on a pivot with weights 50 N and 100 N placed at distances 2 m and 1 m respectively from the pivot. Is the beam in rotational equilibrium?
A · Yes, because clockwise and counterclockwise torques are equal
Torque due to 50 N: \( 50 \times 2 = 100 \text{ Nm} \) counterclockwise.Torque due to 100 N: \( 100 \times 1 = 100 \text{ Nm} \) clockwise.Torques are equal, so beam is in rotational equilibrium.
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A 15 kg box is pulled up a 25° incline with a force of 100 N parallel to the incline. If the coefficient of kinetic friction is 0.2, what is the acceleration of the box? (Use \( g=9.8 \text{ m/s}^2 \))
B · 2.1 m/s²
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Refer to the diagram below showing a free body diagram of a car moving around a banked curve of radius 100 m at speed 20 m/s. The banking angle is 15°. What provides the centripetal force for the car?
A · Horizontal component of the normal force
On a banked curve without friction, the horizontal component of the normal force provides the centripetal force.
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A 10 kg box is pulled by two forces: 30 N east and 40 N north. What is the direction of the resultant acceleration?
A · 53.1° north of east
Direction \( \theta = \tan^{-1} \left( \frac{40}{30} \right) = 53.1° \) north of east.Correction: Option A is correct.
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Which of the following best explains why a passenger feels pushed outward when a car turns sharply to the left?
A · Inertia of the passenger resisting change in motion
The passenger's inertia resists the change in direction, making them feel pushed outward, though the actual force is centripetal inward.
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A 2 kg block is pulled across a rough surface with a force of 20 N at 30° above the horizontal. The coefficient of kinetic friction is 0.1. What is the acceleration of the block? (Use \( g=9.8 \text{ m/s}^2 \))
B · 7.1 m/s²
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Refer to the diagram below showing a free body diagram of a block on a frictionless incline of 45°. The block is released from rest. What is its acceleration down the slope?
A · 6.93 m/s²
Acceleration down slope \( a = g \sin 45° = 9.8 \times 0.707 = 6.93 \text{ m/s}^2 \).
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A block of mass 3.6 kg is placed on a rough horizontal surface with coefficient of kinetic friction 0.28. A force of 18.5 N is applied at an angle of 35° above the horizontal. Find the acceleration of the block.
C · 2.0 m/s²
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Which of the following best defines linear momentum?
A · The product of mass and velocity of an object
Linear momentum is defined as the product of an object's mass and its velocity, representing the quantity of motion.
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Momentum is a vector quantity because it has both magnitude and
B · direction
Momentum depends on velocity, which is a vector, so momentum has both magnitude and direction.
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Refer to the diagram below showing a vector \( \vec{p} \) representing momentum of a particle moving at velocity \( \vec{v} \). Which vector represents the direction of momentum?
A · Vector \( \vec{v} \) (velocity vector)
Momentum vector is in the same direction as the velocity vector since \( \vec{p} = m \vec{v} \).
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What are the SI units of momentum?
A · kg\( \cdot \)m/s
Momentum is mass times velocity, so its units are kilograms times meters per second (kg\( \cdot \)m/s).
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If a body of mass 2 kg moves with a velocity of 3 m/s, what is its momentum?
A · 6 kg\( \cdot \)m/s
Momentum \( p = m \times v = 2 \times 3 = 6 \) kg\( \cdot \)m/s.
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Which of the following expressions correctly represents momentum \( \vec{p} \)?
A · \( \vec{p} = m \vec{v} \)
Momentum is defined as mass multiplied by velocity vector: \( \vec{p} = m \vec{v} \).
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Two objects of masses 3 kg and 2 kg move towards each other with velocities 4 m/s and 3 m/s respectively. What is the total momentum of the system?
A · 6 kg\( \cdot \)m/s
Taking one direction positive, total momentum = \(3 \times 4 + 2 \times (-3) = 12 - 6 = 6\) kg\( \cdot \)m/s.
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Refer to the diagram below showing two carts colliding on a frictionless track. If the total momentum before collision is \( 20 \; \text{kg}\cdot\text{m/s} \), what is the total momentum after collision?
A · 20 kg\( \cdot \)m/s
According to the law of conservation of momentum, total momentum in a closed system remains constant if no external forces act.
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A 5 kg object moving at 10 m/s collides elastically with a stationary 3 kg object. Which principle is used to find their velocities after collision?
A · Conservation of momentum and conservation of kinetic energy
Elastic collisions conserve both momentum and kinetic energy, so both principles are used.
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Two ice skaters push off each other on frictionless ice. Skater A has mass 50 kg and Skater B has mass 70 kg. If Skater A moves away with velocity 3 m/s, what is the velocity of Skater B?
A · -2.14 m/s
By conservation of momentum: \( 50 \times 3 + 70 \times v = 0 \Rightarrow v = -\frac{150}{70} = -2.14 \) m/s (opposite direction).
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Impulse is defined as the change in
A · momentum
Impulse equals the change in momentum of an object when a force is applied over a time interval.
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Refer to the impulse-time graph below. What is the impulse delivered to the object during the time interval from 0 to 4 seconds?
A · 20 N\( \cdot \)s
Impulse is the area under the force-time graph. Here, area = force \( \times \) time = 5 N \( \times \) 4 s = 20 N\( \cdot \)s.
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Which of the following equations correctly relates impulse \( J \) to momentum change?
A · \( J = \Delta p \)
Impulse \( J \) equals the change in momentum \( \Delta p \), i.e., \( J = \Delta p = m v_f - m v_i \).
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A force of 10 N acts on a 2 kg object for 3 seconds. What is the change in momentum of the object?
A · 30 kg\( \cdot \)m/s
Impulse \( = F \times t = 10 \times 3 = 30 \) N\( \cdot \)s, which equals change in momentum.
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In an elastic collision between two objects, which quantities are conserved?
A · Both momentum and kinetic energy
Elastic collisions conserve both total momentum and total kinetic energy of the system.
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Refer to the diagram below showing two spheres colliding inelastically and sticking together. What happens to the kinetic energy after collision?
A · It decreases
In inelastic collisions, kinetic energy is not conserved; some is transformed into other forms like heat or deformation.
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Two objects collide and stick together. If the initial momentum of the system is 15 kg\( \cdot \)m/s, what is the final momentum?
A · 15 kg\( \cdot \)m/s
Momentum is conserved in all collisions, including perfectly inelastic ones where objects stick together.
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In a perfectly elastic head-on collision between two equal masses, what happens to their velocities after collision?
A · They exchange their velocities
In a perfectly elastic collision between equal masses, the two objects exchange their velocities.
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A bullet of mass 0.01 kg moving at 500 m/s embeds into a stationary block of mass 2 kg. What is the velocity of the block-bullet system immediately after the collision?
A · 2.5 m/s
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A car of mass 1000 kg moving at 20 m/s applies brakes and stops in 5 seconds. What is the average force exerted by the brakes?
A · 4000 N opposite to motion
Change in momentum = \( 1000 \times (0 - 20) = -20000 \) kg\( \cdot \)m/s. Force = \( \frac{\Delta p}{\Delta t} = \frac{-20000}{5} = -4000 \) N (opposite to motion).
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Two ice hockey players collide and push off each other. Player A (mass 80 kg) moves backward at 3 m/s, and Player B (mass 60 kg) moves forward. What is the velocity of Player B?
A · 4 m/s forward
By conservation of momentum: \( 80 \times (-3) + 60 \times v = 0 \Rightarrow v = \frac{240}{60} = 4 \) m/s forward.
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Refer to the diagram below showing a system of two masses connected by a spring on a frictionless surface. If the spring suddenly releases and the masses move apart, which principle can be used to find their velocities?
A · Conservation of momentum
Since no external force acts horizontally, total momentum before and after release is conserved.
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