MDCAT Physics chapter 2: Physics · Force and motion MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
27 exam-style questions on Physics · Force and motion, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
Ten Physics · Force and motion questions with answers
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. A boy can throw a ball to a greatest horizontal distance of 40 m on level ground. Throwing with the same speed vertically upwards, the greatest height he can reach is
A
40 m
B
10 m
C
80 m
D
20 m
Show answer
Answer: D. The greatest range is u2/g = 40 m (at 45°). Thrown straight up, the ball rises to u2/2g, half of this: 20 m. 10 m is the height reached in the 45° throw.
2. A fielder moves his hands backwards while catching a fast cricket ball. This
A
makes the change in momentum of the ball smaller than mv
B
increases the impulse that his hands give to the ball
C
shortens the time of impact, which reduces the force
D
spreads the stop over a longer time, reducing the force
Show answer
Answer: D. The ball's change in momentum, and so the impulse, is fixed at mv. Since F = Δp/Δt, making the stopping time longer makes the average force on his hands smaller.
3. A ball is thrown vertically upwards and falls back to the thrower's hand. Taking upwards as positive and ignoring air resistance, its velocity–time graph is
A
a V-shaped line that just touches the time axis at the highest point
B
a horizontal line lying above the time axis
C
a straight line of negative slope crossing the time axis
D
a curve that is steepest at the highest point
Show answer
Answer: C. The acceleration is −g throughout, so the graph is a straight line of slope −g: positive velocity on the way up, zero at the top, negative on the way down. A V-shape would be the graph of speed, not velocity.
4. A 4.0 kg trolley moving at 2.0 m s−1 collides head-on with a 2.0 kg trolley moving towards it at 4.0 m s−1, and the two couple together. The kinetic energy lost in the collision is
A
zero
B
8.0 J
C
24 J
D
16 J
Show answer
Answer: C. Total momentum = 4.0 × 2.0 − 2.0 × 4.0 = 0, so the coupled trolleys stop and all the kinetic energy, ½ × 4.0 × 2.02 + ½ × 2.0 × 4.02 = 8.0 + 16 = 24 J, is lost. Zero total momentum does not mean zero kinetic energy.
5. Two balls are projected from the same point at the same instant with equal speeds, one at 30° and the other at 60° to the horizontal. Ignoring air resistance,
A
both land at the same point, but the 30° ball lands first
B
both land at the same point at the same instant
C
both land at the same point, but the 60° ball lands first
D
the 30° ball lands first and farther away
Show answer
Answer: A. Complementary angles give equal ranges, R = u2 sin 2θ/g. The time of flight 2u sin θ/g is shorter for the smaller angle, so the 30° ball lands first.
6. A body starts from rest and moves with uniform acceleration. It covers 3.0 m in the first second of its motion. The distance it covers in the fourth second alone is
A
12 m
B
21 m
C
48 m
D
9.0 m
Show answer
Answer: B. The distance in the first second is ½a = 3.0 m, so a = 6.0 m s−2. The distances in successive seconds are in the ratio 1 : 3 : 5 : 7, so the fourth second gives 7 × 3.0 = 21 m. 48 m is the total distance in four seconds, not the distance in the fourth second alone.
7. The impulse of a force acting on a body is equal to
A
the change in kinetic energy it produces
B
the change in momentum it produces
C
the force divided by the time it acts for
D
the rate of change of momentum it produces
Show answer
Answer: B. Impulse = FΔt, and since F = Δp/Δt this equals Δp, the change in momentum; its unit N s is the same as kg m s−1. The rate of change of momentum is the force itself.
8. The velocity–time graph of a body is a straight line falling from +10 m s−1 at t = 0 to −10 m s−1 at t = 4.0 s. The distance it travels in these 4.0 s is
A
20 m
B
zero
C
10 m
D
40 m
Show answer
Answer: A. The line crosses the time axis at t = 2.0 s. Each triangle has area ½ × 2.0 × 10 = 10 m, so the distance is 10 + 10 = 20 m. The two areas cancel to give zero displacement, which is not the distance.
9. A body moves 30 m due east in 10 s and then 10 m due west in the next 10 s. Its average velocity for the whole 20 s is
A
1.0 m s−1 east
B
2.0 m s−1 east
C
1.0 m s−1 west
D
2.0 m s−1 west
Show answer
Answer: A. Average velocity = net displacement ÷ time = (30 − 10) m east ÷ 20 s = 1.0 m s−1 east. 2.0 m s−1 is the average speed (40 m of path ÷ 20 s).
10. A projectile is launched with horizontal and vertical velocity components of 30 m s−1 and 40 m s−1. Taking g = 10 m s−2, its horizontal range on level ground is
A
120 m
B
320 m
C
250 m
D
240 m
Show answer
Answer: D. Time of flight T = 2uy/g = 2 × 40/10 = 8.0 s, so R = uxT = 30 × 8.0 = 240 m. 120 m uses only the time to the highest point, and 250 m is the range that the same speed (50 m s−1) would give at 45°.
Physics · Force and motion: questions students ask
How many Physics · Force and motion MCQs are there?
27, each with the answer and a short explanation, across 3 topics: Kinematics, Newton's laws and momentum, Projectile motion. Every chapter test draws ten of them, so you can take it many times.
Is the MDCAT Physics chapter 2 test free?
Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.
How is the chapter test marked?
The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
