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Step-by-Step Solution
Step 1: Identify the known quantities
• Mass of the bullet, $m = 10\,\text{g} = 0.01\,\text{kg}$
• Initial speed of the bullet, $u = 400\,\text{m s}^{-1}$
• Mass of the wood block, $M = 2\,\text{kg}$
• Length of the string, $l = 5\,\text{m}$
• Vertical rise of the block’s center of gravity, $h = 0.10\,\text{m}$
• Final speed of the bullet (after passing through the block), $v = ?$
Step 2: Use conservation of energy for the block
After the bullet passes through the block, the block acquires some speed $v_1$. As the block is suspended, it then rises through a height $h$ before coming momentarily to rest.
By the principle of conservation of energy (assuming no energy loss apart from that transferred to the block by the bullet):
$\frac{1}{2} M v_1^2 = M g h$
Therefore,
$v_1 = \sqrt{2 g h}$.
Substituting $g = 10\,\text{m s}^{-2}$ and $h = 0.10\,\text{m}$:
$v_1 = \sqrt{2 \times 10 \times 0.10} = \sqrt{2}\,\text{m s}^{-1}$.
Step 3: Apply conservation of linear momentum
Consider the horizontal direction just before and just after the collision between the bullet and the block. Initially, the block is at rest, and the bullet is moving with speed $u$.
Total momentum before collision:
$m \times u$.
Total momentum after collision (the block moves with speed $v_1$ and the bullet emerges with speed $v$):
$M \times v_1 + m \times v$.
By conservation of momentum:
$m u = M v_1 + m v$.
Substitute the known values ($m = 0.01\,\text{kg}$, $u = 400\,\text{m s}^{-1}$, $M = 2\,\text{kg}$, and $v_1 = \sqrt{2}\,\text{m s}^{-1}$):
$0.01 \times 400 = 2 \times \sqrt{2} + 0.01 \times v$.
This simplifies to:
$4 = 2\sqrt{2} + 0.01\,v$.
Rearrange to solve for $v$:
$4 - 2\sqrt{2} = 0.01\,v \quad \Longrightarrow \quad v = \frac{4 - 2\sqrt{2}}{0.01}$.
Numerically, $2\sqrt{2} \approx 2 \times 1.414 = 2.828$, so:
$4 - 2.828 = 1.172 \quad \Longrightarrow \quad v \approx \frac{1.172}{0.01} = 117.2\,\text{m s}^{-1}.$
Rounding suitably, $v \approx 120\,\text{m s}^{-1}$.
Step 4: Conclude the bullet’s final speed
Hence, the speed of the bullet after it emerges out of the block is
$120\,\text{m s}^{-1}$.
Relevant Diagram
