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

Calculation of Velocity and Momentum 

Given that the velocity of the block is derived from the energy equation:

\[\text{Velocity of block} = \sqrt{2gh} = \sqrt{2 \times g \times 0.1} = \sqrt{2} \, \text{m/s}\]

Next, applying the principle of conservation of momentum, we have:

\[P_i = P_f\]

Now, let's apply the given values to the momentum equation:

\[(10 \times 10^{-3}) \times 400 + 0 = 2 \times \sqrt{2} + (10 \times 10^{-3}) V\]

Simplifying and solving for \(V\):

\[V = \frac{(4 - 2 \sqrt{2})}{10 \times 10^{-3}} = 120 \, \text{m/s}\]

Answer 2

Given:
- Mass of bullet: $m = 10\,g = 0.01\,kg$ 
- Initial velocity of bullet: $u = 400\,m/s$
- Mass of block: $M = 2\,kg$
- Vertical rise of block: $h = 10\,cm = 0.1\,m$

Step 1: Use conservation of energy for the block's rise
\(\frac{1}{2} M V^2 = M g h \Rightarrow V = \sqrt{2 g h} = \sqrt{2 \cdot 9.8 \cdot 0.1} = \sqrt{1.96} \approx 1.4\,m/s\)
This is the velocity of the block immediately after the collision.

Step 2: Use conservation of momentum for the collision
\(\text{Initial momentum} = m u,\quad \text{Final momentum} = m v + M V\)
\(\Rightarrow m u = m v + M V\)
Substituting known values:
\(0.01 \cdot 400 = 0.01 \cdot v + 2 \cdot 1.4\)
\(4 = 0.01 v + 2.8 \Rightarrow 0.01 v = 1.2 \Rightarrow v = \frac{1.2}{0.01} = \boxed{120\,m/s}\)

Final Answer:
\(\boxed{120\, m/s}\)