JEE Main 2024 — 27 January, Shift 2 — Question 49

Let: $u$ be the initial velocity of the bullet when it enters the target. $a$ be the constant deceleration of the bullet inside the target. We use the kinematic equation: $v^2 = u^2 + 2as$.

\textbf{Part 1: After travelling 4 cm} The bullet travels a distance $s_1 = 4 \text{ cm} = 0.04 \text{ m}$. It loses one-third of its velocity, so its velocity becomes $v_1 = u - \frac{1}{3}u = \frac{2}{3}u$.

Applying the kinematic equation for this part:

$$v_1^2 = u^2 + 2as_1$$ $$\left(\frac{2}{3}u\right)^2 = u^2 + 2a(0.04)$$ $$\frac{4}{9}u^2 = u^2 + 0.08a$$

Rearranging to find $a$ in terms of $u$:

$$0.08a = \frac{4}{9}u^2 - u^2$$ $$0.08a = \left(\frac{4-9}{9}\right)u^2$$ $$0.08a = -\frac{5}{9}u^2$$ $$a = -\frac{5}{9 \times 0.08}u^2$$ $$a = -\frac{5}{0.72}u^2$$ $$a = -\frac{500}{72}u^2$$ $$a = -\frac{125}{18}u^2$$

\textbf{Part 2: Total distance until the bullet comes to rest} Let $s_{total}$ be the total distance the bullet penetrates from its entry point until it comes to rest. Initial velocity = $u$ Final velocity = $0$

Applying the kinematic equation for the entire motion:

$$v_{final}^2 = u^2 + 2as_{total}$$ $$0^2 = u^2 + 2\left(-\frac{125}{18}u^2\right)s_{total}$$ $$0 = u^2 - \frac{250}{18}u^2 s_{total}$$ $$0 = u^2 - \frac{125}{9}u^2 s_{total}$$

Since $u \neq 0$, we can divide the entire equation by $u^2$:

$$0 = 1 - \frac{125}{9} s_{total}$$ $$\frac{125}{9} s_{total} = 1$$ $$s_{total} = \frac{9}{125} \text{ m}$$

\textbf{Part 3: Further distance penetrated} The problem states that the bullet penetrates further $D \times 10^{-3} \text{ m}$ before coming to rest, after already travelling $s_1 = 0.04 \text{ m}$. The further distance penetrated ($s_{further}$) is:

$$s_{further} = s_{total} - s_1$$ $$s_{further} = \frac{9}{125} \text{ m} - 0.04 \text{ m}$$ $$s_{further} = \frac{9}{125} - \frac{4}{100} \text{ m}$$ $$s_{further} = \frac{9}{125} - \frac{1}{25} \text{ m}$$

To perform the subtraction, find a common denominator:

$$s_{further} = \frac{9}{125} - \frac{1 \times 5}{25 \times 5}$$ $$s_{further} = \frac{9}{125} - \frac{5}{125}$$ $$s_{further} = \frac{4}{125} \text{ m}$$

We are given that this further distance is $D \times 10^{-3} \text{ m}$.

$$D \times 10^{-3} = \frac{4}{125}$$

To express $\frac{4}{125}$ in the form $X \times 10^{-3}$, we can multiply the numerator and denominator by 8:

$$\frac{4}{125} = \frac{4 \times 8}{125 \times 8} = \frac{32}{1000}$$ $$\frac{32}{1000} = 32 \times 10^{-3} \text{ m}$$

Comparing $D \times 10^{-3} = 32 \times 10^{-3}$:

$$D = 32$$

The final answer is $\boxed{32}$.