Q165: Question Bank - Physics Class 11 Motion in One Dimension Q165 Solution: If a body loses half of its velocity on penetrating 3 cm in a…
Students often incorrectly assume that penetration depth is directly proportional to velocity, leading them to predict the body will travel another 3 cm. You must recognize that stopping distance is proportional to the square of the initial velocity ($s \propto v^2$); since the velocity is halved, the remaining kinetic energy is one-fourth of the initial, so the additional distance is $\frac{1}{4}$ of the first 3 cm, which is 0.75 cm (or 1 cm in the context of the provided options, though strictly 0.75 cm is the mathematical result for 'more' distance, note that the provided solution calculates $x=1$ which implies a specific interpretation or typo in the standard problem variant where the answer is often 1 cm if the first part was different, but based on the provided solution steps, the trap is assuming linear velocity-distance relationship). Actually, looking at the provided solution's final step: $x=1$ is derived from $\frac{v^2}{4} = \frac{v^2}{4} x \implies x=1$? No, $2 \cdot \frac{v^2}{8} = \frac{v^2}{4}$. So $\frac{v^2}{4} = \frac{v^2}{4} x \implies x=1$. Wait, the solution says $x=1$. Let's re-verify. $v_f^2 = v_i^2 - 2as$. $0 = (v/2)^2 - 2(v^2/8)x$. $0 = v^2/4 - v^2/4 x$. $v^2/4 x = v^2/4$. $x=1$. The math in the solution yields 1 cm. The trap is assuming linear proportionality ($v \propto s$) instead of quadratic ($v^2 \propto s$).
Multiple Choice Options
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