# Aprojectile of mass moving with speed in the direction strikes a stationary target of mass head-on. the collision is elastic.

Aprojectile of mass moving with speed in the direction strikes a stationary target of mass 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: , , and .)

## This Post Has 3 Comments

1. naidaisha18 says:

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

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

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,

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

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

2. jorgefrom584 says:

Compatible and speedy

3. ComicSans10 says:

net primary productivity

Explanation: i just took my test and this was correct!