Question

A metal ball of mass $2kg$ moving with a velocity of $36\raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$h$}\right.$ has a head on collision with a stationary ball of mass $3kg$. If after the collision, the two balls move together, find the loss in kinetic energy due to collision.

Answer 1

Step 1: Given data

Mass $m_1=2kg$

Mass $m_2=3kg$

Velocity $v_1=36\raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$h$}\right.=10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Velocity $v_2=0$

Step 2: Concept used

Law of conservation of momentum,

$\Rightarrow m_1{v}_1+m_2{v}_2=\left(m_1+m_2\right)v$

Step 3: Finding the loss in kinetic energy due to collision

$$\begin{gathered} \Rightarrow m_1{v}_1+m_2{v}_2=\left(m_1+m_2\right)v \\ \Rightarrow v=\frac{m_1{v}_1}{m_1+m_2} \\ \Rightarrow v=\frac{10\times 2}{5}=4\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Now,

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

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

Then,

Loss in $K.E=100-40=60J$

Hence, the loss in kinetic energy due to collision is $60J$.