A 50 g bullet moving with velocity 10 m/s strikes a block of mass 950 g at rest and gets embedded in it. The loss in kinetic energy will be :
Answer & explanation
Correct answer: option 2
Initial K.E. of system = K.E. of the bullet = \(\frac{1}{2}m_s v^2_s\)
Conservation of Linear Momentum : \(m_s v_s + 0 = m_{sys}.v_{sys}\)
\(\Rightarrow v_{sys} = \frac{m_s\times v_s}{m_{sys}}\)
\(\Rightarrow v_{sys} = \frac{50\times 10}{50+950} = 0.5 m/s\)
Fractional loss in KE = \(\frac{\frac{1}{2}m_s v_s^2-\frac{1}{2}m_{sys} v_{sys}^2}{\frac{1}{2}m_s v^2_s}\)
Put \(m_s = 50\times 10^{-3} \text{ kg; } v_s = 10 \text{ m/s; } m_{sys} = 1 \text{ kg and } v_{sys} = 0.5 \text{ m/s} \)
We get :
Fractional loss = \(\frac{95}{100} \Rightarrow 95\)%