Question

A body of mass 40 kg having velocity 4 m/s collides with another body of mass 60 kg having velocity 2 m/s. If the collision is inelastic, then loss in kinetic energy will be

• A. 440 J
• B. 392 J
• C. 48 J
• D. 144 J

Answer 1

The correct option is C. 48 J.

Given,

$m_1=40kg$

$u_1=4m/s$

$m_2=60kg$

$u_2=2m/s$

Since the collision is inelastic, both masses would move together at the velocity v,

According to the conservation of momentum,

$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

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

$\Rightarrow 40\times 4+60\times 2=(40+60)v$

$\Rightarrow 280=100v$

$\Rightarrow v=2.8m/s$

Initial kinetic energy = $K{E}_i=\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=0.5\times 40\times 16+0.5\times 60\times 4=440J$

The final $K{E}_f=\frac{1}{2}(m_1+m_2)v^2=0.5(40+60){2.8}^2=392J$

Thus the difference in the kinetic energy = $440-392=48J$.