Answer:
[tex]y=2.64m[/tex]
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
[tex]p_{i}=p_{f}\\m_{1}v_{1}+m_{2}v_{2}=m_{t}v_{t}[/tex]
So the combined velocity vt is:
[tex]v_{t}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{t}}\\[/tex]
Since the two balls 1 and 2 are moving in opposite direction
So
[tex]v_{t}=\frac{m_{1}v_{1}-m_{2}v_{2}}{m_{t}}\\[/tex]
Substitute the given values
[tex]v_{t}=\frac{(3kg)(20m/s)-(2kg)(12m/s)}{(3+2)kg}\\ v_{t}=7.2 m/s[/tex]
We have the equation for motion with constant acceleration is given by:
[tex]v^2=v_{o}^2+2g(y-y_{o})\\[/tex]
At initial position y₀=0 and vt=v-v₀
So
[tex]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[/tex]
