Question

Two billiard balls each of mass 0.05 \(\text {kg}\) moving in opposite directions with speed 6 \(\text m \,\text s^{-1}\)collide and rebound with the same speed. What is the impulse imparted to each ball due to the other ?

Answer 1

Given Data

• Mass each: \(m = 0.05\) kg
• Initial speeds: 6 m/s (opposite directions)
• Rebound: same speed 6 m/s (opposite directions)
• Head-on elastic collision

Ball 1 →→→ 6 m/s ←←← 6 m/s Ball 2
Collision
Ball 1 ←←← 6 m/s →→→ 6 m/s Ball 2

Impulse Definition

Impulse = change in momentum:

$$\vec{J} = \Delta \vec{p} = m(\vec{v}_f - \vec{v}_i)$$

Calculate for Ball 1

$$v_{1i} = +6 \, \text{m/s}, \quad v_{1f} = -6 \, \text{m/s}$$ $$\Delta v_1 = -6 - (+6) = -12 \, \text{m/s}$$ $$J_1 = m \Delta v_1 = 0.05 \times (-12) = -0.6 \, \text{kg·m/s}$$

For Ball 2 (Symmetric)

$$v_{2i} = -6 \, \text{m/s}, \quad v_{2f} = +6 \, \text{m/s}$$ $$\Delta v_2 = +12 \, \text{m/s}$$ $$J_2 = 0.05 \times 12 = +0.6 \, \text{kg·m/s}$$

Impulse Imparted

Magnitude: 0.6 kg·m/s
(Ball 1: -0.6, Ball 2: +0.6 → equal & opposite)

Verification: Conservation Laws

Ball	p_initial	p_final	Δp (Impulse)
Ball 1	+0.3	-0.3	-0.6
Ball 2	-0.3	+0.3	+0.6
Total	0	0	0 ✓

Physical Insight

• Perfectly elastic: speeds unchanged, directions reversed
• Impulse magnitude same for both (Newton's 3rd law)
• Total momentum conserved (isolated system)
• Contact force × time = momentum change