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

`y=2.64m`

Step-by-step 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)=(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)=(m_(1)v_(1)-m_(2)v_(2))/(m_(t))\\`

Substitute the given values

`v_(t)=((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=(v^2-v_(o)^2)/(2g)\\ y=(v_(t)^2)/(2g)\\ y=((7.2m/s)^2)/(2(9.8m/s^2))\\ y=2.64m`