A bullet from a gun is fired on a rectangular wooden block with velocity $u$. When the bullet travels $24 ext

Question

A bullet from a gun is fired on a rectangular wooden block with velocity $u$. When the bullet travels $24	ext{ cm}$ through the block along its length horizontally, velocity of bullet becomes $u/3$. Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is:

Accepted Answer

1. **Identify the Principle:** The resistive force assumed constant causes a constant deceleration ($a$). We can use the Work-Energy Theorem or the kinematic equation $v^2 - u^2 = 2as$ [Class 11 Physics, Ch 5, Sec 5.6].
2. **Analyze First Part of Motion:**
   - Initial velocity $v_i = u$
   - Final velocity $v_f = u/3$
   - Displacement $s_1 = 24	ext{ cm}$
   $$ (u/3)^2 - u^2 = 2as_1 $$
   $$ \frac{u^2}{9} - u^2 = 2a(24) $$
   $$ -\frac{8u^2}{9} = 48a \implies a = -\frac{u^2}{54} $$
3. **Analyze Total Motion:** Let the total length of the block be $L$. The bullet enters with $u$ and stops ($v=0$) after traversing distance $L$.
   $$ 0^2 - u^2 = 2aL $$
   $$ -u^2 = 2\left(-\frac{u^2}{54}ight)L $$
   $$ 1 = \frac{2L}{54} \implies L = 27	ext{ cm} $$
4. **Alternative Method:** Calculate the remaining distance $s_2$ from velocity $u/3$ to $0$.
   $$ 0^2 - (u/3)^2 = 2as_2 $$
   $$ -\frac{u^2}{9} = 2\left(-\frac{u^2}{54}ight)s_2 \implies s_2 = 3	ext{ cm} $$
   Total length $L = 24 + 3 = 27	ext{ cm}$.

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