Class XI · Chapter 04
Laws of Motion — Exam Practice
Practice questions in CBSE, NEET, and JEE Main style covering Newton’s laws, momentum conservation, and friction.
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.
1CBSE Board Exam Style
Q1. Why does a gun recoil when a bullet is fired from it?
Answer
By conservation of momentum, the total momentum of the gun-bullet system before firing (zero) must equal the total after firing. The bullet’s forward momentum is exactly balanced by the gun’s equal and opposite backward momentum — the recoil.
By conservation of momentum, the total momentum of the gun-bullet system before firing (zero) must equal the total after firing. The bullet’s forward momentum is exactly balanced by the gun’s equal and opposite backward momentum — the recoil.
Q2. Starting from Newton’s second law in the form F = dp/dt, derive the familiar relation F = ma for a body of constant mass.
Answer
F = dp/dt = d(mv)/dt. For constant mass, m can be taken outside the derivative:
F = m(dv/dt) = ma, since dv/dt is by definition the acceleration a.
F = dp/dt = d(mv)/dt. For constant mass, m can be taken outside the derivative:
F = m(dv/dt) = ma, since dv/dt is by definition the acceleration a.
This shows F = ma is really a special case of the more general momentum-based statement, valid only when mass doesn’t change.
Q3. (a) State the law of conservation of linear momentum, and derive it using Newton’s third law for a two-body collision. (b) A bullet of mass 20 g moving at 300 m/s strikes a stationary 5 kg block and embeds itself in it. Find the common velocity of the block and bullet after impact.
Answer
(a) The total momentum of an isolated system remains constant if no external force acts on it. For two colliding bodies exerting equal and opposite forces on each other (Newton’s third law) over the same contact time, the resulting equal-and-opposite impulses mean their combined momentum before and after collision is unchanged.
(a) The total momentum of an isolated system remains constant if no external force acts on it. For two colliding bodies exerting equal and opposite forces on each other (Newton’s third law) over the same contact time, the resulting equal-and-opposite impulses mean their combined momentum before and after collision is unchanged.
(b) By conservation of momentum: m₁v₁ = (m₁+m₂)v.
0.02 × 300 = (0.02 + 5)v → 6 = 5.02v → v ≈ 1.2 m/s.
2NEET Style (Single-Correct MCQ)
Q4. The coefficient of static friction between two surfaces is generally:
Solution
μₛ > μₖ typically — which is why it takes more force to start something sliding than to keep it sliding.
μₛ > μₖ typically — which is why it takes more force to start something sliding than to keep it sliding.
Q5. A block slides down a frictionless incline of angle θ. Its acceleration along the incline is:
Solution
Only the component of gravity along the incline (mg sinθ) accelerates the block; the perpendicular component is balanced by the normal force.
Only the component of gravity along the incline (mg sinθ) accelerates the block; the perpendicular component is balanced by the normal force.
Q6. Newton’s First Law of Motion is essentially a statement defining:
Solution
It defines both the concept of inertia (resistance to change in motion) and, implicitly, what an inertial reference frame is — one in which the law actually holds.
It defines both the concept of inertia (resistance to change in motion) and, implicitly, what an inertial reference frame is — one in which the law actually holds.
Q7. Impulse delivered to an object equals its:
Solution
J = FΔt = Δp — impulse is exactly the change in momentum, useful especially when force varies or acts briefly.
J = FΔt = Δp — impulse is exactly the change in momentum, useful especially when force varies or acts briefly.
3JEE Main Style
Q8. A force of 20 N acts on a 4 kg mass initially at rest, for 3 s. Find its final velocity.
Solution
a = F/m = 20/4 = 5 m/s². v = u + at = 0 + 5×3 = 15 m/s.
a = F/m = 20/4 = 5 m/s². v = u + at = 0 + 5×3 = 15 m/s.
Q9. Rocket propulsion, even in the vacuum of space, is best explained by:
Solution
The rocket expels exhaust gas backward; conservation of momentum for the rocket-plus-exhaust system requires the rocket to gain forward momentum — no external medium to “push against” is needed.
The rocket expels exhaust gas backward; conservation of momentum for the rocket-plus-exhaust system requires the rocket to gain forward momentum — no external medium to “push against” is needed.
Q10. A 2 kg block rests on a rough horizontal surface with μₛ = 0.3. A horizontal force of 5 N is applied. Find the friction force acting, and state whether the block moves. (g = 10 m/s²)
Solution
Limiting static friction: fmax = μₛmg = 0.3 × 2 × 10 = 6 N.
Limiting static friction: fmax = μₛmg = 0.3 × 2 × 10 = 6 N.
Since the applied force (5 N) is less than fmax (6 N), the block does not move, and static friction exactly matches the applied force: f = 5 N.
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