A bullet of mass 10 g moving horizontally with a velocity of 400 m s ^ - 1 strikes a wooden block of mass 2 kg which is suspended by light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be

A bullet of mass 10 g moving horizontally with a velocity of 400 m/s strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the center of gravity of the block rises by a vertical distance of 10 cm. The speed of the bullet after it emerges from the block will be:

(a) 100 m/s

(b) 80 m/s

(c) 120 m/s

(d) 160 m/s

Detailed Answer:

We are given: - Mass of the bullet \( m_b = 10 \, \text{g} = 0.01 \, \text{kg} \), - Initial velocity of the bullet \( u_b = 400 \, \text{m/s} \), - Mass of the block \( m_{\text{block}} = 2 \, \text{kg} \), - Length of the string \( L = 5 \, \text{m} \), - Vertical displacement of the block's center of gravity \( h = 10 \, \text{cm} = 0.1 \, \text{m} \). First, we calculate the potential energy gained by the block using the formula for gravitational potential energy: \[ PE = m_{\text{block}} g h. \] Substituting the values: \[ PE = 2 \times 10 \times 0.1 = 2 \, \text{J}. \] This potential energy is gained from the kinetic energy of the bullet after the collision. Now, applying the principle of conservation of momentum for the collision (assuming the bullet moves horizontally and there is no external force in the horizontal direction): \[ m_b u_b = m_b v_b + m_{\text{block}} v_{\text{block}}. \] Where: - \( u_b \) is the initial velocity of the bullet, - \( v_b \) is the final velocity of the bullet, - \( v_{\text{block}} \) is the final velocity of the block after the collision. After the collision, the block rises to a height \( h \), so the velocity of the block just after the collision is found using the energy principle: \[ \frac{1}{2} m_{\text{block}} v_{\text{block}}^2 = PE. \] Substituting the value of \( PE \): \[ \frac{1}{2} \times 2 \times v_{\text{block}}^2 = 2 \quad \Rightarrow \quad v_{\text{block}}^2 = 2 \quad \Rightarrow \quad v_{\text{block}} = \sqrt{2} \approx 1.41 \, \text{m/s}. \] Now, using conservation of momentum: \[ 0.01 \times 400 = 0.01 \times v_b + 2 \times 1.41. \] Simplifying: \[ 4 = 0.01 v_b + 2.82 \quad \Rightarrow \quad 0.01 v_b = 4 - 2.82 = 1.18. \] Solving for \( v_b \): \[ v_b = \frac{1.18}{0.01} = 118 \, \text{m/s}. \] Therefore, the speed of the bullet after emerging from the block is approximately: \[ \boxed{120 \, \text{m/s}} \ (\text{Option (c)}). \]