Question

A block of mass 1.2 kg moving at a speed of 20 cm/s collides head-on with a similar block kept at rest. The coefficient of restitution is $\frac{3}{5}$. Find the loss of kinetic energy during the collision.

• A. 7.7$\times$ ${10}^{-2}$ J
• B. 7.7$\times$ ${10}^{-1}$ J
• C. 7.7$\times$ ${10}^{-3}$ J
• D. 7.7$\times$ ${10}^{-4}$ J

Answer 1

The correct option is C

7.7$\times$ ${10}^{-3}$ J

Suppose the first block moves at a speed $v_1$ and the second at $v_2$ after the collision. Since the collision is

head-on, the two blocks move along the original direction of motion of the first block.

By conservation of linear momentum,

(1.2 kg)(20 cm/s) = (1.2 kg)$v_1$ + (1.2 kg)$v_2$

or, $v_1+v_2$ = 20 cm/s ...........(i)

The velocity of separation is $v_2-v_1$ and the velocity of approach is 20 cm/s. as the coefficient of restitution is $\frac{3}{5}$, we have,

$v_2-v_1$ = $\frac{3}{5}\times 20$ cm/s = 12 cm/s ...........(ii)

by (i) and (ii)

$v_1$ = 4 cm/s and $v_2$ = 16 cm/s

the loss in kinetic energy is

$\frac{1}{2}$$\left(1.2kg\right)\left[\right(20cm/s{)}^2-(4cm/s{)}^2-(16cm/s{)}^2]$

= (0.6 kg)[0.04 $m^2/s^2$ - 0.0016 $m^2/s^2$ - 0.0256 $m^2/s^2$]

= (0.6 kg)(0.0128 $m^2/s^2$) = 7.7$\times$ ${10}^{-3}$ J.