Discussion Forum

A metal ball of mass 2 kg moving with a velocity of 36 km/h has a head on collision with a stationary ball of mass 3 kg. If  after the collision , the two balls move together , the loss in kinetic energy due to collision is 

Answer:

Explanation:

 Apply conservation of momentum

 m₁v₁ = (m₁+m₂)v

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

 Here, v₁ = 36 km/hr=10 m/s

 m₁ =2 kg , m₂=3 kg

 $v= \frac{10\times2}{5}=4 m/s$

 K.E ( initial ) $= \frac{1}{2}\times2\times(10)^{2}=100 J$

K.E (final) $= \frac{1}{2}\times(3+2)\times(4)^{2}=40 J$

 Loss in K.E =100-40=60

Althernatively use this formula

$-\triangle E_{k}=\frac{1}{2}\frac{m_{1}m_{2}}{(m_{1}+m_{2})}(u_{1}-u_{2})^{2}$