JEE Advanced · Physics · 4. Motion in 2D

A small block is connected to one end of a massless spring of un-stretched length $4.9 \mathrm{~m}$. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by $0.2 \mathrm{~m}$ and released from rest at $t=0$. It then executes simple harmonic motion with angular frequency $\omega=\pi / 3 \mathrm{rad} / \mathrm{s}$. Simultaneously at $t=0$, a small pebble is projected with speed $v$ form point $P$ at an angle of $45^{\circ}$ as shown in the figure. Point $P$ is at a horizontal distance of $10 \mathrm{~m}$ from $O$. If the pebble hits the block at $t=1 \mathrm{~s}$, the value of $v$ is (take $g$ $=10 \mathrm{~m} / \mathrm{s}^{2}$ )

A $\sqrt{50} \mathrm{~m} / \mathrm{s}$
B $\sqrt{51} \mathrm{~m} / \mathrm{s}$
C $\sqrt{52} \mathrm{~m} / \mathrm{s}$
D $\sqrt{53} \mathrm{~m} / \mathrm{s}$

Verified Solution

Answer & Solution

Correct Answer

(A) $\sqrt{50} \mathrm{~m} / \mathrm{s}$

Step-by-step Solution

Detailed explanation

Time of flight of projectile,

$T=\frac{2 V \sin \theta}{g}$

$\therefore \quad 1=\frac{2 V \sin 45^{\circ}}{g} \quad \therefore \quad V=\sqrt{50} \mathrm{~ms}^{-1}$

Hence pebble is projected with a speed $V=\sqrt{50} \mathrm{~ms}^{-1}$

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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