Question

A bullet of mass $10\,g$ moving horizontally with a velocity of $400 \, ms^{-1}$ strikes a wooden block of mass $2 \,kg$ which is suspended by a 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. $100 \, ms^{-1}$
• B. $80 \, ms^{-1}$
• C. $120 \, ms^{-1}$
• D. $160 \, ms^{-1}$

Answer 1

The Correct Option is C

To solve this problem, we will use the principle of conservation of momentum and energy considerations. Let's break it down step by step:

1. First, we need to understand the scenario where a bullet of mass 10 \, g = 0.01 \, kg strikes a wooden block of mass 2 \, kg.
2. The initial velocity of the bullet is 400 \, ms^{-1}.
3. After the bullet passes through, the block (with bullet embedded at initial collision) swings to a height of 10 \, cm = 0.1 \, m.
4. Using the conservation of energy for the block's swing, we have:
   • Potential energy at the height: mgh = 2 \times 9.8 \times 0.1 = 1.96 \, J.
   • This was converted from kinetic energy when the block first started to swing:
   • \frac{1}{2} \times 2 \times v_{block}^2 = 1.96 \, J
   • Solving for v_{block} gives:
   • v_{block} = \sqrt{\frac{1.96}{1}} = \sqrt{1.96} \approx 1.4 \, ms^{-1}.
5. Next, apply the principle of conservation of linear momentum before and after the collision:
   • Initial momentum: 0.01 \times 400 = 4 \, kg \cdot ms^{-1}.
   • Final momentum: 0.01 \times v_{bullet} + 2 \times v_{block}.
   • Using the calculated block velocity, substitute into the conservation equation:
   • 4 = 0.01 \times v_{bullet} + 2 \times 1.4
   • Simplifying, we get: 4 = 0.01 \times v_{bullet} + 2.8
   • 0.01 \times v_{bullet} = 4 - 2.8 = 1.2
   • v_{bullet} = \frac{1.2}{0.01} = 120 \, ms^{-1}.
6. Thus, the speed of the bullet after it emerges from the block is 120 \, ms^{-1}.