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

The correct option is 48 J.

Let us analyze the given parameters of the collision:
Mass of the first body, $m_1=40\text{kg}$
Initial velocity of the first body, $u_1=4\text{m/s}$
Mass of the second body, $m_2=60\text{kg}$
Initial velocity of the second body, $u_2=2\text{m/s}$

Since the collision is completely inelastic (implied by the maximum loss of kinetic energy where the two bodies stick together after collision), they move with a common velocity $v$ after the collision.

By the law of conservation of linear momentum:
$m_1{u}_1+m_2{u}_2=(m_1+m_2)v$

Substituting the values:
$(40\times 4)+(60\times 2)=(40+60)v$
$160+120=100v$
$280=100v$
$v=2.8\text{m/s}$

Now, let's calculate the initial kinetic energy ($K_i$):
$K_i=\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2$
$K_i=\frac{1}{2}\left(40\right)\left(4^2\right)+\frac{1}{2}\left(60\right)\left(2^2\right)$
$K_i=20\left(16\right)+30\left(4\right)$
$K_i=320+120=440\text{J}$

Let's calculate the final kinetic energy ($K_f$):
$K_f=\frac{1}{2}(m_1+m_2)v^2$
$K_f=\frac{1}{2}\left(100\right)\left({2.8}^2\right)$
$K_f=50\left(7.84\right)=392\text{J}$

Therefore, the loss in kinetic energy ($\mathrm{\Delta K}$) is:
$\mathrm{\Delta K}=K_i-K_f$
$\mathrm{\Delta K}=440\text{J}-392\text{J}=48\text{J}$