Collisions and Momentum: Numerical Problems Set (3)

So $v_2 - v_1 = 2.5$.

Add $4 \times$ (second eq) to first: $v_1 + 4 v_2 + 4(v_2 - v_1) = 5 + 10$, i.e. $-3 v_1 + 8 v_2 = 15$.

Easier: from second, $v_1 = v_2 - 2.5$. Substitute: $(v_2 - 2.5) + 4 v_2 = 5 \Rightarrow 5 v_2 = 7.5 \Rightarrow v_2 = 1.5$ m/s. Then $v_1 = -1$ m/s.

$KE_i = \tfrac{1}{2}(1)(25) = 12.5$ J. $KE_f = \tfrac{1}{2}(1)(1) + \tfrac{1}{2}(4)(2.25) = 0.5 + 4.5 = 5$ J.

Final answer: $v_1 = -1$ m/s (rebounds), $v_2 = 1.5$ m/s, energy loss $= 60\%$.

Why This Works

Momentum is always conserved in collisions (no external horizontal force). The coefficient of restitution captures the elasticity of the collision — $e = 1$ is perfectly elastic, $e = 0$ is perfectly inelastic, anything in between is partial.

Solving simultaneously gives both post-collision velocities. The energy loss in any non-elastic collision goes into heat, sound, and deformation.

Alternative Method

Use the standard formulas:

Plug in: $v_1 = (1 - 2)/5 \times 5 = -1$ m/s. $v_2 = 1.5/5 \times 5 = 1.5$ m/s. Same answer in 30 seconds — worth memorising for JEE Main.