Two blocks M₁ and M₂ of equal mass m can move freely on a horizontal frictionless surface. Block M₂ carries a

NCERT Exemplar · Q21

M_1

(a) While the spring is fully compressed, all the kinetic energy of

M_1

is stored as potential energy of the spring.

(b) While the spring is fully compressed, the momentum of the system is not conserved, even though the final momentum equals the initial momentum.

M_1

is a state of rest.

(d) If the surface on which the blocks move has friction, then the collision cannot be elastic.

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Concept understanding — Conservation of Momentum in Collisions

Conservation of Momentum in Collisions

When two objects collide — two billiard balls, two train carriages coupling together, a bat striking a ball — each one exerts a force on the other for the brief instant they are in contact. This concept looks specifically at what those internal, mutual forces do to the total momentum of the colliding pair.

Setting Up the Idea

Consider two bodies of masses $m_1$ and $m_2$ , moving with velocities $\vec{u}_1$ and $\vec{u}_2$ just before they collide, and velocities $\vec{v}_1$ and $\vec{v}_2$ just after. During the collision, body 1 pushes on body 2 with some force $\vec{F}$ , and by Newton's third law, body 2 pushes back on body 1 with force $-\vec{F}$ — equal in size, opposite in direction, acting for exactly the same contact time $\Delta t$ .

Since impulse equals change in momentum ( $\vec{F}\Delta t = \Delta \vec{p}$ ):

Change in momentum of body 1: $-\vec{F}\Delta t$
Change in momentum of body 2: $+\vec{F}\Delta t$

Add the two changes together and they cancel to zero. Whatever momentum one body loses, the other gains — the total is untouched.

Important

For any collision where no external force acts on the pair (an isolated system), the total linear momentum just before the collision equals the total linear momentum just after:

$m_1 \vec{u}_1 + m_2 \vec{u}_2 = m_1 \vec{v}_1 + m_2 \vec{v}_2$

Why "Internal" Forces Don't Change the Total

The collision force between the two bodies is an internal force of the two-body system — it acts between the objects, not from outside. Internal forces always come in Newton's-third-law pairs, and any equal-and-opposite pair produces zero net impulse on the system as a whole. External forces — gravity, friction from the floor, air resistance — are usually negligible during a collision because contact lasts only a fraction of a second, far too short for a small external force to produce a noticeable impulse.

Watch out

Momentum conservation applies to the system's total, not to each object individually. One ball can slow down drastically and the other speed up — their individual momenta change freely; only the sum stays fixed.

Worked Example

A 2 kg ball moving at 3 m/s (rightward) strikes a stationary 1 kg ball. After the collision the 2 kg ball moves at 1 m/s in the same direction. Find the velocity of the 1 kg ball.

Take rightward as positive.

p_{\text{before}} = (2)(3) + (1)(0) = 6\ \text{kg m/s}

Let $v$ be the final velocity of the 1 kg ball:

p_{\text{after}} = (2)(1) + (1)(v) = 2 + v

Setting these equal: $2 + v = 6 \implies v = 4\ \text{m/s}$ . …

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At maximum compression both blocks move together at $v/2$ , so they still have kinetic energy — not all of $M_1$ 's KE is in the spring (A wrong). Momentum is conserved at every instant (B wrong). For equal masses the spring eventually pushes $M_1$ to rest and $M_2$ off with $v$ (C correct), and surface friction would dissipate energy, preventing an elastic collision (D correct).

(A) — False

At the instant of maximum compression both blocks move with a common velocity. By momentum conservation:

mv = (m+m)v_c \implies v_c = v/2

So the system still has kinetic energy $\tfrac12(2m)(v/2)^2 = \tfrac14 mv^2$ . The spring PE at that instant is only $\tfrac12 mv^2 - \tfrac14 mv^2 = \tfrac14 mv^2$ — not all of $M_1$ 's initial KE.

(B) — False

The surface is frictionless and the spring forces are internal, so there is no external horizontal force. Momentum is conserved at every instant, including full compression.

(C) — True …

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(A) — False

(B) — False

(C) — True …

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