A bullet is fired into a fixed target. It loses (1/3)rd of its velocity after travelling 4 cm. It penetrates further p × 10^–3 m before coming to rest. Find p.

The correct answer is 32.

Let us analyze the motion of the bullet inside the target step-by-step using the equations of motion under uniform retardation.

Step 1: Understand the initial motion
Let the initial velocity of the bullet before entering the target be $v_0$.
The bullet travels a distance $s_1=4\text{cm}$ and loses $\frac{1}{3}$ of its initial velocity.

Thus, the velocity of the bullet after travelling $4\text{cm}$ is:
$v_1=v_0-\frac{1}{3}{v}_0=\frac{2}{3}$
Using third equation of motion, $v^2-u^2=2as$, we have:
$v_1^2-v_0^2=2a{s}_1$
Substituting the values:
${\left(\frac{2}{3}\right)}^2-v_0^2=2a\left(4\right)$
$\frac{4}{9}$
$-\frac{5}{9}$
$a=-\frac{5}{72}$
where $a$ is the constant retardation inside the target.

Step 2: Find the further penetration distance
Let the bullet penetrate a further distance $s_2$ before coming to rest.
For this part of the motion, the initial velocity is $v_1=\frac{2}{3}$ and the final velocity is $v_2=0$.
Applying the equation of motion again:
$v_2^2-v_1^2=2a{s}_2$
$0-{\left(\frac{2}{3}\right)}^2=2\left(-\frac{5}{72}\right)s_2$
$-\frac{4}{9}$
$s_2=\frac{4}{9}$

Step 3: Convert the distance and find p
We are given that the further penetration distance is $p\times {10}^{-3}\text{m}$.
Converting $s_2$ to meters:
$s_2=3.2\text{cm}=3.2\times {10}^{-2}\text{m}=32\times {10}^{-3}\text{m}$
Comparing this with the given term:
$p\times {10}^{-3}=32\times {10}^{-3}$
Thus, $p=32$.