Question

A projectile of mass m1 moving with speed v1 in the +x direction strikes a stationary target of mass m2 head-on. The collision is elastic. Use the Momentum Principle and the Energy Principle to determine the final velocities of the projectile and target, making no approximations concerning the masses. (Use the following as necessary: m1, m2, and v1.)

Answer 1

Answer:

$v''=\frac{2m_{1}v_{1}}{m_{1}+m_{2}}$

$v' = \frac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1}$

Explanation:

first mass, m1

second mass, m2

initial velocity of m1 is v1.

initial velocity of m2 is zero.

Let the final velocity of m1 is v' and the final velocity of m2 is v''.

Collision is elastic so the coefficient of restitution is 1.

By using the conservation of momentum

m1 x v1 + m2 x 0 = m1 x v + m2 x v''

m1 v1 = m1 v' + m2 v'' .... (1)

by the formula of coefficient of restitution

$1=\frac{v''-v'}{v_{1}-0}$

v1 = v'' - v' .... (2)

So, v' = v'' - v1 put in equation (1)

m1 v1 = m1 x (v'' - v1) + m2 v''

m1 v1 = m1 v'' - m1 v1 + m2 v''

2 m1 v1 = (m1 + m2) v''

$v''=\frac{2m_{1}v_{1}}{m_{1}+m_{2}}$

so,

$v' = \frac{2m_{1}v_{1}}{m_{1}+m_{2}}-v_{1}$

$v' = \frac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1}$