Question

Two bodies A and B of masses 1.5 kg and 3 kg are moving with velocities 20 m/s and 15 m/s respectively. If the same retarding force is applied on the two bodies, then the ratio of the distances travelled by the bodies A and B before they come to rest is

• A. 1:1
• B. 8:9
• C. 2:3
• D. 3:8

Hint:

Work-energy theorem: Work done = change in kinetic energy.

Answer 1

The Correct Option is B

Let $m_A = 1.5$ kg, $v_A = 20$ m/s, $m_B = 3$ kg, and $v_B = 15$ m/s. Let the retarding force be $F$. Using the work-energy theorem, the work done by the retarding force is equal to the change in kinetic energy: $FS_A = \frac{1}{2}m_A v_A^2$ $FS_B = \frac{1}{2}m_B v_B^2$ $\frac{S_A}{S_B} = \frac{\frac{1}{2}m_A v_A^2}{\frac{1}{2}m_B v_B^2} = \frac{m_A v_A^2}{m_B v_B^2} = \frac{(1.5)(20^2)}{(3)(15^2)} = \frac{1.5 \times 400}{3 \times 225} = \frac{600}{675} = \frac{8}{9}$.