Chapter 9, Force and Laws of Motion, is primarily based on formulas from class 9, with a bit of conceptual material. With good practice, students will be able to solve numericals in exams with ease. We have included 24 questions below with varying levels of difficulty.
Formulas
- Momentum: p = m × v
- Change in momentum: Δp = m × (v − u)
- Acceleration: a = (v − u) ÷ t
- Force (Newton’s second law): F = m × a
- Impulse: Impulse = F × t = Δp
- Conservation of momentum (1-D): m1u1 + m2u2 = m1v1 + m2v2
- For uniform acceleration (when needed): v² = u² + 2 × a × s
- Average speed (uniform acceleration): average speed = (u + v) ÷ 2
- Unit conversion: 1 km/h = 5/18 m/s
Class 9 Science (NCERT) – Chapter 8: Force and Laws of Motion
Basic
Q1) Find the momentum of a body of mass 2 kg moving with a velocity of 4 m/s.
Solution:
Given: m = 2 kg, v = 4 m/s
Use p = m × v
p = 2 × 4 = 8 kg m/s
Answer: 8 kg m/s
Q2) A force produces an acceleration of 1 m/s² in an object of mass 1 kg. Find the force.
Solution:
Given: m = 1 kg, a = 1 m/s²
Use F = m × a
F = 1 × 1 = 1 N
Answer: 1 N
Q3) A 5 kg object changes its velocity from 2 m/s to 6 m/s. Find the change in momentum.
Solution:
Given: m = 5 kg, u = 2 m/s, v = 6 m/s
Use Δp = m × (v − u)
Δp = 5 × (6 − 2)
Δp = 5 × 4 = 20 kg m/s
Answer: 20 kg m/s
Q4) A constant force of 10 N acts for 2 s on a body. Find the impulse.
Solution:
Given: F = 10 N, t = 2 s
Use Impulse = F × t
Impulse = 10 × 2 = 20 N s
Answer: 20 N s
Q5) A body’s velocity changes from 0 m/s to 12 m/s in 3 s. Find its acceleration.
Solution:
Given: u = 0 m/s, v = 12 m/s, t = 3 s
Use a = (v − u) ÷ t
a = (12 − 0) ÷ 3 = 4 m/s²
Answer: 4 m/s²
Q6) An object of mass 2 kg slides with constant velocity 4 m/s on a frictionless horizontal table. Find the force required to keep it moving with the same velocity.
Solution:
Given: constant velocity
Constant velocity means acceleration a = 0
Use F = m × a
F = 2 × 0 = 0 N
Answer: 0 N
Standard
Q7) A bullet of mass 10 g travelling at 150 m/s comes to rest in 0.03 s. Find (i) acceleration, (ii) force on the bullet.
Solution:
Given: m = 10 g = 0.01 kg, u = 150 m/s, v = 0 m/s, t = 0.03 s
(i) Use a = (v − u) ÷ t
a = (0 − 150) ÷ 0.03
a = −5000 m/s²
(ii) Use F = m × a
F = 0.01 × (−5000) = −50 N
Answer: Acceleration = −5000 m/s², Force = −50 N
Q8) An object of mass 100 kg is accelerated from 5 m/s to 8 m/s in 6 s. Find (i) initial momentum, (ii) final momentum, (iii) force.
Solution:
Given: m = 100 kg, u = 5 m/s, v = 8 m/s, t = 6 s
(i) Initial momentum p1 = m × u
p1 = 100 × 5 = 500 kg m/s
(ii) Final momentum p2 = m × v
p2 = 100 × 8 = 800 kg m/s
(iii) a = (v − u) ÷ t
a = (8 − 5) ÷ 6 = 0.5 m/s²
Force F = m × a
F = 100 × 0.5 = 50 N
Answer: p1 = 500 kg m/s, p2 = 800 kg m/s, F = 50 N
Q9) A 1 kg object moving at 10 m/s sticks to a stationary 5 kg block. Find (i) total momentum before impact, (ii) common velocity after impact.
Solution:
Given: m1 = 1 kg, u1 = 10 m/s, m2 = 5 kg, u2 = 0 m/s
(i) Momentum before = m1u1 + m2u2
= (1 × 10) + (5 × 0)
= 10 kg m/s
(ii) After sticking: total mass = 1 + 5 = 6 kg
Use conservation of momentum:
10 = 6 × v
v = 10 ÷ 6 = 1.67 m/s
Answer: Momentum before = 10 kg m/s, Common velocity = 1.67 m/s
Q10) A dumb-bell of mass 10 kg falls from a height 80 cm. Take g = 10 m/s². Find the momentum just before it hits the floor.
Solution:
Given: m = 10 kg, h = 80 cm = 0.8 m, u = 0, g = 10 m/s²
Use v² = u² + 2 × g × h
v² = 0 + 2 × 10 × 0.8
v² = 16
v = 4 m/s
Momentum p = m × v
p = 10 × 4 = 40 kg m/s
Answer: 40 kg m/s
Q11) A hammer of mass 500 g moving at 50 m/s is stopped in 0.01 s. Find the force of the nail on the hammer.
Solution:
Given: m = 500 g = 0.5 kg, u = 50 m/s, v = 0 m/s, t = 0.01 s
a = (v − u) ÷ t
a = (0 − 50) ÷ 0.01
a = −5000 m/s²
F = m × a
F = 0.5 × (−5000) = −2500 N
Answer: −2500 N
Q12) A motorcar of mass 1200 kg slows from 90 km/h to 18 km/h in 4 s. Find (i) acceleration, (ii) change in momentum, (iii) force.
Solution:
Given: m = 1200 kg, u = 90 km/h, v = 18 km/h, t = 4 s
Convert to m/s:
u = 90 × (5/18) = 25 m/s
v = 18 × (5/18) = 5 m/s
(i) a = (v − u) ÷ t
a = (5 − 25) ÷ 4
a = −5 m/s²
(ii) Δp = m × (v − u)
Δp = 1200 × (5 − 25)
Δp = 1200 × (−20) = −24000 kg m/s
(iii) F = m × a
F = 1200 × (−5) = −6000 N
Answer: a = −5 m/s², Δp = −24000 kg m/s, F = −6000 N
Advance
Q13) The speed of sound in air is 346 m/s and in oxygen is 316 m/s. Find how much more distance sound travels in air than in oxygen in the same time of 2 s.
Solution:
Given: v(air) = 346 m/s, v(oxygen) = 316 m/s, time = 2 s
Distance in air = 346 × 2 = 692 m
Distance in oxygen = 316 × 2 = 632 m
Difference = 692 − 632 = 60 m
Answer: 60 m
Q14) A ball of mass 20 g has initial velocity 20 cm/s and comes to rest in 10 s due to friction. Find (i) acceleration, (ii) frictional force.
Solution:
Given: m = 20 g = 0.02 kg, u = 20 cm/s = 0.2 m/s, v = 0, t = 10 s
(i) a = (v − u) ÷ t
a = (0 − 0.2) ÷ 10
a = −0.02 m/s²
(ii) F = m × a
F = 0.02 × (−0.02)
F = −0.0004 N
Answer: a = −0.02 m/s², F = −0.0004 N
Q15) A bullet of mass 10 g strikes a sand-bag at 1000 m/s and gets embedded after travelling 5 cm. Find (i) acceleration, (ii) resistive force, (iii) time taken to stop.
Solution:
Given: m = 10 g = 0.01 kg, u = 1000 m/s, v = 0, s = 5 cm = 0.05 m
(i) Use v² = u² + 2 × a × s
0 = (1000)² + 2 × a × 0.05
0 = 1000000 + 0.1a
a = −10000000 m/s²
(ii) F = m × a
F = 0.01 × (−10000000)
F = −100000 N
(iii) a = (v − u) ÷ t
−10000000 = (0 − 1000) ÷ t
t = 1000 ÷ 10000000
t = 0.0001 s
Answer: a = −10000000 m/s², F = −100000 N, t = 0.0001 s
Q16) A force of 5 N produces acceleration 8 m/s² on mass m1 and 24 m/s² on mass m2. If both masses are tied together, find the acceleration produced by the same 5 N force.
Solution:
Given: F = 5 N
m1 = F ÷ a1
m1 = 5 ÷ 8 = 0.625 kg
m2 = F ÷ a2
m2 = 5 ÷ 24 = 0.2083 kg
Total mass = 0.625 + 0.2083 = 0.8333 kg
Acceleration a = F ÷ (total mass)
a = 5 ÷ 0.8333 = 6 m/s²
Answer: 6 m/s²
Q17) Two persons push a motorcar of mass 1200 kg at uniform velocity. With three persons, it gets acceleration 0.2 m/s². If each pushes with the same force, find the force applied by one person.
Solution:
Let force by one person = F
With two persons and uniform velocity:
Net force = 0
So resistive force = 2F
With three persons:
Net force = 3F − 2F = F
But net force = m × a
F = 1200 × 0.2
F = 240 N
Answer: 240 N
Q18) Using Q17 result, if 4 persons push the same car with the same effort, find the acceleration.
Solution:
From Q17: one person’s force F = 240 N
Resistive force = 2F = 2 × 240 = 480 N
Force by 4 persons = 4F = 4 × 240 = 960 N
Net force = 960 − 480 = 480 N
Acceleration a = net force ÷ m
a = 480 ÷ 1200
a = 0.4 m/s²
Answer: 0.4 m/s²
HOTS
Q19) In Q9, find the loss of kinetic energy due to collision. Use common velocity v = 1.67 m/s.
Solution:
Given: m1 = 1 kg, u1 = 10 m/s, m2 = 5 kg, u2 = 0, v = 1.67 m/s
Initial KE = (1/2) × m1 × u1² + (1/2) × m2 × u2²
Initial KE = (1/2) × 1 × 10² + 0
Initial KE = 50 J
Final KE = (1/2) × (m1 + m2) × v²
Final KE = (1/2) × 6 × (1.67)²
Final KE = 3 × 2.7889
Final KE = 8.3667 J
Loss of KE = 50 − 8.3667
Loss of KE = 41.6333 J
Answer: 41.63 J
Q20) For the car in Q12, verify that impulse equals change in momentum.
Solution:
From Q12: m = 1200 kg, u = 25 m/s, v = 5 m/s, t = 4 s
Change in momentum:
Δp = m × (v − u)
Δp = 1200 × (5 − 25)
Δp = −24000 kg m/s
Force from Q12:
F = −6000 N
Impulse = F × t
Impulse = −6000 × 4
Impulse = −24000 N s
Answer: Δp = −24000 kg m/s and Impulse = −24000 N s (equal)
Q21) A hammer of mass 0.5 kg moving at 50 m/s stops in 0.01 s. Find (i) change in momentum, (ii) impulse, (iii) average force.
Solution:
Given: m = 0.5 kg, u = 50 m/s, v = 0, t = 0.01 s
(i) Δp = m × (v − u)
Δp = 0.5 × (0 − 50)
Δp = −25 kg m/s
(ii) Impulse = Δp
Impulse = −25 N s
(iii) Average force = Impulse ÷ t
Average force = −25 ÷ 0.01
Average force = −2500 N
Answer: Δp = −25 kg m/s, Impulse = −25 N s, Force = −2500 N
Q22) In Q12, find the distance covered by the car while slowing down (assume uniform acceleration).
Solution:
From Q12: u = 25 m/s, v = 5 m/s, t = 4 s
Average speed = (u + v) ÷ 2
Average speed = (25 + 5) ÷ 2
Average speed = 15 m/s
Distance = average speed × time
Distance = 15 × 4
Distance = 60 m
Answer: 60 m
Q23) A bullet of mass 10 g travelling at 150 m/s comes to rest in 0.03 s. Find the distance it travels while stopping (assume uniform deceleration).
Solution:
Given: u = 150 m/s, v = 0, t = 0.03 s
Average speed = (u + v) ÷ 2
Average speed = (150 + 0) ÷ 2
Average speed = 75 m/s
Distance = average speed × time
Distance = 75 × 0.03
Distance = 2.25 m
Answer: 2.25 m
Q24) A truck of mass M is moved by force F. It is then loaded with an object equal to its own mass (new mass = 2M) and the driving force is halved (new force = F/2). Find how the acceleration changes.
Solution:
Initially:
a1 = F ÷ M
After loading and halving force:
a2 = (F/2) ÷ (2M)
a2 = F ÷ (4M)
Compare:
a2 ÷ a1 = [F ÷ (4M)] ÷ [F ÷ M]
a2 ÷ a1 = 1/4
Answer: New acceleration becomes 1/4 of the original acceleration.
