Collisions and Momentum: Diagram-Based Questions (1)

Question

A $4$ kg block moving at $5$ m/s collides head-on with a stationary $1$ kg block. The collision is perfectly inelastic. Find the common velocity after collision and the kinetic energy lost.

The diagram shows two blocks on a frictionless surface. Block A (4 kg) moves right with velocity 5 m/s. Block B (1 kg) is at rest. After collision, both stick together and move with a common velocity $v$.

Solution — Step by Step

Why This Works

Momentum conservation works in every collision because the internal forces between the blocks are equal and opposite (Newton’s third law), so they cancel out for the system. Only external forces change total momentum, and on a frictionless horizontal surface, there are no horizontal external forces.

Kinetic energy is not conserved in inelastic collisions — some KE goes into heat, sound, deformation, internal vibration. In a perfectly inelastic collision, the maximum possible KE is lost (subject to momentum conservation).

Alternative Method

Using the reduced-mass formula directly:

$\Delta KE = \tfrac{1}{2}\mu(u_1 - u_2)^2 = \tfrac{1}{2} \cdot \frac{4 \cdot 1}{4 + 1} \cdot (5 - 0)^2 = \tfrac{1}{2}(0.8)(25) = 10 \text{ J} \checkmark$

Common Mistake

Students try to use kinetic energy conservation in inelastic collisions and end up with two equations and one variable, contradiction. KE is conserved only in elastic collisions. For inelastic, use momentum alone, plus the constraint that the two bodies have a common velocity (perfectly inelastic) or a coefficient of restitution $e$ between $0$ and $1$ (partially inelastic).

The keyword check:

• “Stick together” / “perfectly inelastic” → momentum alone, common $v$.
• “Coefficient of restitution given” → momentum + $e = (v_2' - v_1')/(u_1 - u_2)$.
• “Elastic collision” → momentum + KE.