JEE Physics Collisions Notes Set 02

Collisions: Basic Ideas

A collision is a short interaction in which the colliding bodies exert large forces on each other for a very short time. During it the momentum of the system (no external impulse) is conserved: \( m_1\vec u_1+m_2\vec u_2=m_1\vec v_1+m_2\vec v_2 \).

Coefficient of Restitution

\( e=\frac{\text{speed of separation}}{\text{speed of approach}} \) along the line of impact, with 0 <= e <= 1. e = 1 for elastic, e = 0 for perfectly inelastic (bodies move together), 0 < e < 1 for inelastic collisions.

One-Dimensional Collisions

Elastic, head-on: \( v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2 \) and \( v_2=\frac{2m_1}{m_1+m_2}u_1+\frac{m_2-m_1}{m_1+m_2}u_2 \). Equal masses exchange velocities. Perfectly inelastic: \( v=\frac{m_1u_1+m_2u_2}{m_1+m_2} \), with loss of kinetic energy \( \frac12\frac{m_1m_2}{m_1+m_2}(u_1-u_2)^2 \).

Oblique Collisions and Bouncing

For a smooth surface the component of velocity along the surface is unchanged and the normal component becomes e times its value, reversed. A ball dropped from height h rebounds to \( e^2h \).

Explosions

An explosion is a collision in reverse: momentum is conserved but kinetic energy increases. For two fragments from rest, \( m_1v_1=m_2v_2 \) and the energy released is \( \tfrac12m_1v_1^2+\tfrac12m_2v_2^2 \).

Worked Example

A 2 kg body moving at 6 m/s hits a 4 kg body at rest, perfectly inelastically. Find the common speed and the loss of energy.
Solution: \( v=\frac{2\times6}{6}=2\ \text{m/s} \). Loss = \( \tfrac12(2)(36)-\tfrac12(6)(4)=36-12=24\ \text{J} \).