JEE Physics Centre of Mass Notes Set 02

Collisions and Coefficient of Restitution

In a collision the interaction is brief and the impulsive internal forces are far larger than any external force, so momentum is conserved during it. The line along which the impulse acts is the common normal. The coefficient of restitution is defined along the normal as
\[ e=\frac{\text{speed of separation}}{\text{speed of approach}},\qquad 0\le e\le1 \]
For e = 1 (perfectly elastic) kinetic energy is conserved; for e = 0 (perfectly inelastic) the bodies move together afterward; for 0 < e < 1 kinetic energy is lost. Relative speed of separation never exceeds relative speed of approach.

Head-On Collisions

For masses \( m_1,m_2 \) with initial velocities \( u_1,u_2 \) along a line, momentum conservation and the definition of e give
\[ v_1=\frac{(m_1-em_2)u_1+m_2(1+e)u_2}{m_1+m_2},\quad v_2=\frac{(m_2-em_1)u_2+m_1(1+e)u_1}{m_1+m_2} \]
Special cases of an elastic collision (e = 1): equal masses exchange velocities; a light body hitting a very heavy one at rest rebounds with the same speed; a very heavy body hitting a light one at rest leaves it moving at about 2v. For a ball hitting a moving wall the ball's speed after impact increases by twice the wall's speed.
Illustration: a 3 kg ball at 4 m/s hits a 1 kg ball at rest elastically and the velocities become 2 m/s and 6 m/s. A sphere of mass m hitting an identical sphere at rest with restitution e leaves with speeds in the ratio \( \frac{1-e}{1+e} \).

Deformation and Reformation

A collision has two phases. In deformation the bodies are compressed until both have the common velocity \( v=\frac{m_1u_1+m_2u_2}{m_1+m_2} \). The impulse of deformation is \( J_D=m_1(v-u_1) \), and the energy stored is the loss of kinetic energy up to this point, \( U=\frac12\mu v_{rel}^2 \) with reduced mass \( \mu=\frac{m_1m_2}{m_1+m_2} \). In reformation part of this energy is returned, with impulse \( J_R=eJ_D \).
Illustration: a 2 kg body at 4 m/s meets a 4 kg body at 2 m/s head-on; the common velocity is zero, so \( J_D=8 \) N s, \( U=24 \) J, and with the 2 kg body returning at 2 m/s the value of e is 0.5 and \( J_R=4 \) N s.
In a perfectly inelastic collision the loss of kinetic energy is \( \frac12\mu v_{rel}^2 \), the largest possible. For a bullet embedded in a stationary block, the fraction lost is \( \frac{M}{M+m} \).

Oblique Collisions

Resolve velocities along the common normal and the common tangent. Along the normal the pair of impulses changes the momenta and the restitution rule applies. Along the tangent no impulse acts, so each body keeps its tangential velocity (for constant mass). Total momentum is conserved in every direction.
For a ball hitting a smooth floor at angle of incidence \( \alpha \) with speed \( v_0 \): the tangential component \( v_0\sin\alpha \) stays, the normal component becomes \( ev_0\cos\alpha \), so the rebound speed is \( \sqrt{(v_0\sin\alpha)^2+(ev_0\cos\alpha)^2} \) and \( \tan\beta=\frac{\tan\alpha}e \). For e = 1 the angles are equal.
If two equal masses collide elastically and one is initially at rest, they move off at right angles. If they leave at equal speeds with 60 degrees between them, then each has speed \( \frac{V}{\sqrt3} \), \( e=\frac13 \) and the lost energy is \( \frac16mV^2 \).

Bouncing Ball

A ball dropped from height h hits the ground at \( \sqrt{2gh} \) and leaves at \( e\sqrt{2gh} \), so the first rebound height is \( e^2h \) and the impulse from the ground is \( mu(1+e) \) for impact speed u. If the ball rebounds to h/4, then \( e=\frac12 \); from 9 m with e = 2/3 it rises 4 m.

Variable Mass and Rockets

When mass is added or ejected at a rate \( \mu=\left|\frac{dm}{dt}\right| \) with relative velocity \( v_{rel} \), an extra thrust of magnitude \( v_{rel}\mu \) acts on the main body. The thrust acts along the direction of the ejected mass's relative velocity when mass is gained, and opposite to it when mass is lost. Solve in four steps: list the real forces, add the thrust, find the net force, and use \( F_{net}=m\frac{dv}{dt} \) with the instantaneous mass.
For a rocket expelling gas at relative speed \( v_r \), thrust \( =v_r\left|\frac{dm}{dt}\right| \) upward and \( v=u-gt+v_r\ln\frac{m_0}m \). In free space this is \( v=v_r\ln\frac{m_0}m \). With \( m_0=1000 \) kg, 10 kg/s burn, \( v_r=2000 \) m/s and 60 s, \( v\approx1233 \) m/s. For constant acceleration a, \( m=m_0e^{-at/u} \).
Cases with sand: when sand falls onto a moving flat car at a constant rate \( \mu \), the sand has no horizontal velocity, so \( v=\frac{Ft}{m_0+\mu t} \). When sand spills out through a hole in the floor, it leaves with the car's velocity, there is no thrust, and \( a=\frac F{m_0-\mu t} \).