JEE PYQ: Motion in a Straight Line - Question ID 1ad94d5642e1 (JEE Main 2022)

Core Formula & Concept:

This problem involves the motion of a bullet under a constant resistive force inside a wooden block. The key physics concepts and formulas are:

• Work-Energy Theorem: When a constant force $F$ acts on a body, the work done by the force equals the change in kinetic energy of the body. $W = \Delta K = K_f - K_i$ Here, the resistive force $F$ does negative work (since it opposes motion), so: $-F \cdot s = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$ where $s$ is the displacement, $m$ is the mass of the bullet, and $v_i$, $v_f$ are initial and final velocities.
• Constant Resistance Implies Constant Deceleration: Since the resistive force is constant, the deceleration $a$ is also constant. We can use kinematic equations for uniformly decelerated motion: $v_f^2 = v_i^2 + 2 a s$ Here, $a$ is negative (deceleration), so: $v_f^2 = v_i^2 - 2 |a| s$

We are given: - After penetrating $s_1 = 4 \text{ cm}$, velocity becomes $v_1 = \frac{v_0}{3}$, where $v_0$ is the initial velocity. - The bullet stops after travelling a total distance $s_{\text{total}} = 4 + x \text{ cm}$. We need to find $x$.

--- Step-by-Step Derivation:

Step 1: Apply Work-Energy Theorem for the first 4 cm

Let $m$ = mass of bullet, $v_0$ = initial velocity, $v_1 = \frac{v_0}{3}$ = velocity after 4 cm. Work done by resistive force $F$ over distance $s_1 = 4 \text{ cm}$: $-F \cdot s_1 = \frac{1}{2} m v_1^2 - \frac{1}{2} m v_0^2$ Substitute $v_1 = \frac{v_0}{3}$: $-F \cdot 4 = \frac{1}{2} m \left(\frac{v_0}{3}\right)^2 - \frac{1}{2} m v_0^2$ $-4F = \frac{1}{2} m \left(\frac{v_0^2}{9} - v_0^2\right)$ $-4F = \frac{1}{2} m v_0^2 \left(\frac{1}{9} - 1\right)$ $-4F = \frac{1}{2} m v_0^2 \left(-\frac{8}{9}\right)$ $-4F = -\frac{4}{9} m v_0^2$ Multiply both sides by $-1$: $4F = \frac{4}{9} m v_0^2$ Divide both sides by 4: $F = \frac{1}{9} m v_0^2 \quad \text{(Equation 1)}$

Step 2: Apply Work-Energy Theorem until bullet stops

Total distance travelled before stopping: $s_{\text{total}} = 4 + x$ Final velocity $v_f = 0$ Work done by resistive force over total distance: $-F \cdot s_{\text{total}} = \frac{1}{2} m \cdot 0^2 - \frac{1}{2} m v_0^2$ $-F (4 + x) = -\frac{1}{2} m v_0^2$ Multiply both sides by $-1$: $F (4 + x) = \frac{1}{2} m v_0^2 \quad \text{(Equation 2)}$

Step 3: Substitute $F$ from Equation 1 into Equation 2

From Equation 1: $F = \frac{1}{9} m v_0^2$ Substitute into Equation 2: $\frac{1}{9} m v_0^2 (4 + x) = \frac{1}{2} m v_0^2$ Cancel $m v_0^2$ from both sides (assuming $m \neq 0$, $v_0 \neq 0$): $\frac{1}{9} (4 + x) = \frac{1}{2}$ Multiply both sides by 9: $4 + x = \frac{9}{2}$ $4 + x = 4.5$ Subtract 4: $x = 0.5$

Conclusion: The value of $x$ is $0.5$ cm, which corresponds to option C.

--- Common Traps & Exam Tip:

Trap 1: Misapplying Kinematic Equations Without Considering Signs
Students often forget that deceleration is negative acceleration. If they use $v_f^2 = v_i^2 + 2 a s$ without accounting for the negative sign of $a$, they get incorrect results. Always define deceleration as $-|a|$.

Trap 2: Incorrect Use of Work-Energy Theorem
Some students forget that the work done by a resistive force is negative. They write $F \cdot s = \Delta K$ instead of $-F \cdot s = \Delta K$. This leads to sign errors and wrong answers.

Trap 3: Assuming Linear Velocity-Distance Relationship
Students sometimes assume velocity decreases linearly with distance, which is incorrect under constant deceleration. Velocity decreases quadratically with distance ($v^2 \propto s$), not linearly.

Exam Tip:
Always start by writing down known quantities and the physics principle to apply. In this case, since force is constant, the work-energy theorem is the most straightforward approach. Avoid unnecessary use of kinematic equations unless explicitly required.