First slide

The steps in collision

Question

A body of mass $5 \times 10^{3}$ kg moving with speed $2 {ms}^{- 1}$ collides with a stationary body of mass $15 \times 10^{3}$ kg inelastically and sticks to it. Then, loss in kinetic energy of the system will be [AIIMS 2019]

Moderate

A

7.5 kJ
B

15 kJ
C

10 kJ
D

5 kJ

Solution

Given, mass of body, $m_{1} = 5 \times 10^{3} kg$
Velocity, $v_{1} = 2 {ms}^{- 1}$
and mass of another body, $m_{2} = 15 \times 10^{3} kg$
For perfectly inelastic collision, e = 0.
$∴$ Loss in kinetic energy of system,
$Δ E_{K} = \frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} \times v_{1}^{2}$
$= \frac{1}{2} \times \frac{5 \times 10^{3} \times 15 \times 10^{3}}{5 \times 10^{3} + 15 \times 10^{3}} \times (2)^{2}$
$= 7.5 \times 10^{3} J$
= 7.5 kJ

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