Answer to Question #176118 in Mechanics | Relativity for Akinola

Question

A snooker ball X of mass 200g moving at 20ms^-1 to hit another stationary ball T of mass 400g. If after the collision Y moves off at 60^o to the initial direction of X while X moves off at 15^o to it's initial direction with a velocity of 12ms^-1,. Calculate the final velocity of T

Answer 1

To be given in question

Befor collision

Ball X

m₁=200gram =0.2kg

u₁=20m/sec 

Ball T

m₂=400gram=0.4kg

u₂=0m/sec 

After collision

 v1=12m/sec

$\theta$ $1=60° ;\theta2=15°$

To be asked in question

V₂=?

We take x-axis components

Momentum conservation law

Before collision=after collision

m₁u₁+m₂u₂=m₁v₁cos$\theta$₁ +m₂v₂cos$\theta$₂

Put value

$=0.2\times20+ 0.4\times0=0.2\times12\times cos60°+0.4\times V( 2)\times cos15°$

=4+0=2.4$\times$ cos60°$+V(2)\times0.4\times cos15°$

4=1.2+V₂$\times0.4 \times cos15°$

2.8=0.386 $\times$ V₂

V₂=7.25m/sec