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CBSE · Class 9 · Science

How Forces Affect Motion

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
A force acting on a mass of 9 kg gives it an acceleration of 6 m/s2. The force is:
  1. 1.5 N
  2. 15 N
  3. 108 N
  4. 54 N
2.
A force of 180 N acts on a body of mass 20 kg. Its acceleration is:
  1. 9 m/s2
  2. 3600 m/s2
  3. 20 m/s2
  4. 11 m/s2
3.
The momentum of a 32 kg object moving at 2 m/s is:
  1. 16 kg m/s
  2. 34 kg m/s
  3. 65 kg m/s
  4. 64 kg m/s
4.
An 8 kg body speeds up from 8 m/s to 10 m/s. The change in its momentum is:
  1. 16 kg m/s
  2. 80 kg m/s
  3. 144 kg m/s
  4. 2 kg m/s
5.
The tendency of an object to resist a change in its state of rest or motion is called:
  1. inertia
  2. force
  3. acceleration
  4. momentum
6.
A 24 kg body's velocity changes from 5 m/s to 53 m/s in 8 s. The force acting on it is:
  1. 168 N
  2. 1152 N
  3. 1272 N
  4. 144 N
7.
A bullet of mass 15 g is fired at 500 m/s from a gun of mass 6 kg. The recoil speed of the gun is:
  1. 1.25 m/s
  2. 2.25 m/s
  3. 12.5 m/s
  4. 1,250 m/s
8.
Assertion (A): A force of 36 N gives a mass of 6 kg an acceleration of 6 m/s2.
Reason (R): By Newton's second law of motion, F = ma.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
A car of mass 1,600 kg speeds up from 5 m/s to 13 m/s in 9 s. Find the force on it. (Give the answer to 2 decimal places where needed.)
10.
A trolley of mass 2 kg moving at 20 m/s hits a 7 kg trolley at rest, and they move off together. Find their common velocity. (Give the answer to 2 decimal places where needed.)
11.
A vehicle of mass 2,200 kg moving at 20 m/s stops in 3 s. Find the retarding force. (Give the answer to 2 decimal places where needed.)
12.
A ball of mass 180 g hits a wall at 21 m/s and bounces straight back at 17 m/s. Find the change in its momentum. (Give the answer to 2 decimal places where needed.)
13.
The same force acts on two bodies of masses 14 kg and 18 kg. Find the ratio of their accelerations.
14.
A force of 126 N gives a body an acceleration of 7 m/s2. Find its mass.
15.
A 1 kg ball moving at 15 m/s hits a 2 kg ball moving the same way at 4 m/s. After the collision the first ball moves at 1 m/s. Find the speed of the second ball.
16.
A force of 34 N acts on a 17 kg body moving at 7 m/s for 5 s. Find its final velocity.
Section C — Word problems4 × 4 = 16 marks
17.
A fielder catches a cricket ball of mass 167 g moving at 18 m/s, stopping it in 0.4 s. Find the average force on the fielder's hands. (Give the answer to 2 decimal places where needed.)
18.
A hockey ball of mass 250 g travelling at 15 m/s is hit back along its path at 7 m/s. Find the change in its momentum. (Give the answer to 2 decimal places where needed.)
19.
A truck of mass 8 tonnes starts from rest and covers 380 m in 19 s with uniform acceleration. Find its acceleration and the force acting on it. (Give the answer to 2 decimal places where needed.)
20.
A rifle of mass 5 kg fires a bullet of mass 35 g at 600 m/s. Find the recoil velocity of the rifle. (Give the answer to 2 decimal places where needed.)
AmbifulWorksheets
Practice that never repeats
CBSE · Class 9 · Science

Answer key

How Forces Affect Motion · AW-9S-FOR-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) 54 NF = ma = 9 × 6
2(a) 9 m/s2a = F ÷ m = 180 ÷ 20
3(d) 64 kg m/sp = mv = 32 × 2
4(a) 16 kg m/sm(v − u) = 8 × 2
5(a) inertia—
6(d) 144 NF = m(v − u) ÷ t = 24 × 48 ÷ 8
7(a) 1.25 m/s0.015 × 500 = 6 × v; v = 1.25
8(a) Both A and R are true, and R is the correct explanation of A.A is true: F = ma = 6 × 6 = 36 N. R is true and explains A.
91,422.22 NF = m(v − u) ÷ t = 1600 × 8 ÷ 9
104.44 m/smomentum is conserved: 2 × 20 = (2 + 7) × v
1114,666.67 NF = m × (0 − 20) ÷ 3; its size is 14,666.67 N
126.84 kg m/sthe direction reverses: 0.18 × (21 + 17)
139 : 7a = F/m, so a1 : a2 = 18 : 14
1418 kgm = F ÷ a = 126 ÷ 7
1511 m/s1 × 15 + 2 × 4 = 1 × 1 + 2 × v
1617 m/sa = 34 ÷ 17 = 2 m/s2; v = 7 + 2 × 5
177.52 NF = m v ÷ t = 0.167 × 18 ÷ 0.4
185.5 kg m/sthe velocity reverses: 0.25 × (15 + 7)
192.11 m/s2; 16,842.11 Ns = ½at2, so a = 2 × 380 ÷ 192; F = 8000 × a
204.2 m/stotal momentum stays zero: 5 × v = 0.035 × 600