Question

An object of mass 0.4 kg moving with velocity of 4 m/s collides with another object of same mass moving in same direction with velocity of 2 m/s. If the collision is perfectly inelastic (plastic), then what will be the loss of K. E due to impact?

Answer 1

$\text{Step 1: Apply momentum conservation}$ $\text{[Ref. Fig. ]}$

Let $v$ be the velocity of both after collision.

Since there is no external force acting on the system, hence momentum is conserved

$P_i=P_f$

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

$\Rightarrow 0.4\times 4+0.4\times 2=(0.4+0.4)v$

$\Rightarrow v=\frac{1.6+0.8}{0.8}=3m/s$

$\text{Step 2: Kinetic Energy}$

Before collision: $K{E}_i=\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2$$=\frac{1}{2}\times 0.4\times (4{)}^2+\frac{1}{2}\times 0.4\times (2{)}^2=4J$

After collision: $K{E}_f=\frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}\times (0.4+0.4)\times (3{)}^2=3.6J$

$\text{Step 3: Loss of Kinetic Energy}$

Loss of kinetic energy due to impact $=K{E}_i-K{E}_f$ $=4-3.6=0.4J$