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CBSE · Class 11 · Physics

Laws of 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 of 48 N acts on a 8 kg body. Its acceleration is:
  1. 0.17 m/s2
  2. 384 m/s2
  3. 7 m/s2
  4. 6 m/s2
2.
A ball of mass 1.9 kg hits a wall at 22 m/s and rebounds at 28 m/s. The magnitude of the change in its momentum is:
  1. 95 kg m/s
  2. 11.4 kg m/s
  3. 41.8 kg m/s
  4. 190 kg m/s
3.
A gun of mass 5 kg fires a bullet of mass 5 g at 728 m/s. The recoil speed of the gun is:
  1. 0.0728 m/s
  2. 7.28 m/s
  3. 728 m/s
  4. 0.728 m/s
4.
A force of 41 N pulls a 5 kg block along a floor where the coefficient of kinetic friction is 0.8. The acceleration of the block is: (Take g = 10 m/s2.)
  1. 0.2 m/s2
  2. 8.2 m/s2
  3. 16.2 m/s2
  4. 1.2 m/s2
5.
Newton's first law of motion defines:
  1. inertia
  2. impulse
  3. power
  4. momentum
6.
A 52 kg person stands on a weighing scale in a lift. When the lift accelerates downwards at 3 m/s2, the scale reads: (Take g = 10 m/s2.)
  1. 156 N
  2. 520 N
  3. 676 N
  4. 364 N
7.
Masses of 9 kg and 2 kg hang over a light frictionless pulley. Their acceleration is: (Take g = 10 m/s2.)
  1. 6.36 m/s2
  2. 10 m/s2
  3. 15.71 m/s2
  4. 8.18 m/s2
8.
Assertion (A): A 4 kg gun firing a 46 g bullet at 691 m/s recoils at 7.9465 m/s.
Reason (R): The gun and the bullet move off with equal speeds.
  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 550 kg moving at 20 m/s is stopped in 8 s. Find the average braking force.
10.
A force of 16 N acts on a ball of mass 1 kg for 0.13 s. Find the impulse and the change in the ball's velocity.
11.
Two blocks of 9 kg and 8 kg are in contact on a frictionless table. A force of 60 N pushes the 9 kg block. Find the acceleration and the force between the blocks.
12.
In an Atwood machine, masses of 7 kg and 2 kg hang over a light frictionless pulley. Find the acceleration and the tension in the string. (Take g = 10 m/s2.)
13.
A 72 kg person stands on a scale in a lift. Find the reading when the lift (a) moves up at constant speed, (b) accelerates up at 4 m/s2, (c) accelerates down at 4 m/s2. (Take g = 10 m/s2.)
14.
A 14 kg box rests on a floor (μs = 0.7, μk = 0.5). Find the least horizontal force to start it moving. If a force of 116 N is applied, what is its acceleration? (Take g = 10 m/s2.)
15.
A car takes a level circular turn of radius 200 m. If the greatest safe speed is 23 m/s, find the coefficient of friction between the tyres and the road. (Take g = 10 m/s2.)
16.
A machine gun fires 18 bullets a second, each of mass 43 g, at 225 m/s. Find the force needed to hold the gun in place.
Section C — Word problems4 × 4 = 16 marks
17.
Three blocks of 10 kg, 4 kg and 3 kg are joined by light strings on a frictionless table. A force of 16 N pulls the 3 kg block (the last one). Find the acceleration and the tension in each string.
18.
A 18 kg block slides down an incline at 37° to the horizontal with no friction. Find the normal reaction, and the acceleration. (Take g = 10 m/s2, sin 37° = 3/5, cos 37° = 4/5.)
19.
A cricket ball of mass 0.11 kg moving at 37 m/s is caught and stopped. Find the average force on the hands if the catch takes (a) 0.05 s, (b) 0.37 s. Why do fielders pull their hands back?
20.
A 70 kg person stands on a weighing scale in a lift that starts from rest, accelerates up at 1 m/s2 for 5 s, moves at constant speed for 5 s, then decelerates at 1 m/s2 to rest. Find the scale reading in each part and the greatest speed of the lift. (Take g = 10 m/s2.)
AmbifulWorksheets
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CBSE · Class 11 · Physics

Answer key

Laws of Motion · AW-11P-LOM-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) 6 m/s2a = F/m
2(a) 95 kg m/sthe velocity reverses: m(u1 + u2)
3(d) 0.728 m/sM V = m v: 5V = 0.005 × 728
4(a) 0.2 m/s2friction = μmg = 40 N
5(a) inertia—
6(d) 364 NR = m(g − a)
7(a) 6.36 m/s2a = (m1 − m2)g/(m1 + m2)
8(c) A is true, but R is false.A is true: 4 × V = 0.046 × 691, so V = 7.9465 m/s. R is false.
91,375 NF = m(v − u)/t
102.08 N s; 2.08 m/simpulse = Ft = mΔv
113.53 m/s2; 28.24 Na = F/(m1 + m2); contact force = m2a
125.56 m/s2; 31.11 Na = (m1 − m2)g/(m1 + m2); T = 2m1m2g/(m1 + m2)
13(a) 720 N; (b) 1,008 N; (c) 432 NR = m(g ± a)
1498 N; 3.29 m/s2limiting friction μs mg; once moving, friction μk mg
150.2645μmg = mv2/r, so μ = v2/(rg)
16174.15 Nforce = rate of change of momentum = n m v
17a = 0.94 m/s2; tension between the first two blocks = 9.41 N; between the second and third = 13.18 Neach string pulls everything behind it
18N = 144 N; a = 6 m/s2along the slope: mg sin θ − μmg cos θ = ma
19(a) 81.4 N; (b) 11 N. Pulling back increases the stopping time, so the same change in momentum needs a smaller force.F = change in momentum ÷ time
20770 N, then 700 N, then 630 N; greatest speed 5 m/sR = m(g + a) going up faster; R = m(g − a) while slowing