Question

Two bodies A and B of masses 20 kg and 5 kg respectively are at rest. Due to the action of a force of 40 N separately, if the two bodies acquire equal kinetic energies in times \( t_A \) and \( t_B \) respectively, then \( t_A : t_B = \)

• A. 1:2
• B. 2:1
• C. 2:5
• D. 5:6

Hint:

Kinetic Energy \( KE = \frac{1}{2}mv^2 \). If a constant force F acts on a body of mass m starting from rest, its velocity after time t is \( v=at = (F/m)t \). So, \( KE = \frac{1}{2}m(Ft/m)^2 = \frac{F^2 t^2}{2m} \). If \(KE_A = KE_B\) and \(F\) is the same, then \( \frac{t_A^2}{m_A} = \frac{t_B^2}{m_B} \implies \frac{t_A}{t_B} = \sqrt{\frac{m_A}{m_B}} \).

Answer 1

The Correct Option is B

Let \( m_A = 20 \) kg and \( m_B = 5 \) kg.
Force \( F = 40 \) N.
The bodies start from rest.
Acceleration of body A: \( a_A = F/m_A = 40/20 = 2 \, \text{m/s}^2 \).
Acceleration of body B: \( a_B = F/m_B = 40/5 = 8 \, \text{m/s}^2 \).
Velocity after time \(t\): \( v = at \) (since initial velocity is 0).
Kinetic energy (KE) = \( \frac{1}{2}mv^2 = \frac{1}{2}m(at)^2 = \frac{1}{2}ma^2t^2 \).
Also, \( a = F/m \), so \( KE = \frac{1}{2}m\left(\frac{F}{m}\right)^2 t^2 = \frac{1}{2}m \frac{F^2}{m^2} t^2 = \frac{F^2 t^2}{2m} \).
Given that the kinetic energies acquired are equal: \( KE_A = KE_B \).
\[ \frac{F^2 t_A^2}{2m_A} = \frac{F^2 t_B^2}{2m_B} \] Since F is the same for both: \[ \frac{t_A^2}{m_A} = \frac{t_B^2}{m_B} \] \[ \frac{t_A^2}{t_B^2} = \frac{m_A}{m_B} \] \[ \left(\frac{t_A}{t_B}\right)^2 = \frac{20}{5} = 4 \] \[ \frac{t_A}{t_B} = \sqrt{4} = 2 \] So, \( t_A : t_B = 2:1 \).
This matches option (2).