Question

A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis.Just before the collision,one ball,of mass 3.0 kg,is moving up- ward at 20 m/s and the other ball, of mass 2.0 kg, is moving down- ward at 12 m/s. How high do the combined two balls of putty rise above the collision point

Answer 1

Answer:

$y=2.64m$

Explanation:

Given data

Ball one

mass m₁=3.0kg

velocity v₁=20 m/s

Ball second

mass m₂=2.0 kg

velocity v₂=12 m/s

First we need the speed of combined ball.Since the system conserves the linear momentum

$p_{i}=p_{f}\\m_{1}v_{1}+m_{2}v_{2}=m_{t}v_{t}$

So the combined velocity vt is:

$v_{t}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{t}}$

Since the two balls 1 and 2 are moving in opposite direction

So

$v_{t}=\frac{m_{1}v_{1}-m_{2}v_{2}}{m_{t}}$

Substitute the given values

$v_{t}=\frac{(3kg)(20m/s)-(2kg)(12m/s)}{(3+2)kg}\\ v_{t}=7.2 m/s$

We have the equation for motion with constant acceleration is given by:

$v^2=v_{o}^2+2g(y-y_{o})$

At initial position y₀=0 and vt=v-v₀

So

$v^{2}=v_{o}^2+2g(y-0)\\ y=\frac{v^2-v_{o}^2}{2g}\\ y=\frac{v_{t}^2}{2g}\\ y=\frac{(7.2m/s)^2}{2(9.8m/s^2)}\\ y=2.64m$