A wedge Y with mass of 10 kg and all frictionless surfaces and the… | JEE Main 2026 (5 Apr, Shift 1)

Q5JEE Main 2026 · 5 APR · SHIFT 1Difficult

Connected Bodies & Pulleys

A wedge $Y$ with mass of 10 kg and all frictionless surfaces and the inclined surface making $37^\circ$ with horizontal. A block $X$ with mass 2 kg is placed at the highest point of the wedge as shown in figure is at rest. At $t = 0$ wedge ( $Y$ ) is pulled toward right with constant force $(f)$ of 24 N. Taking the block $X$ at rest at $t = 0$ , the time taken by it to slide down 8.8 m on the slope, while $Y$ is on the move, is ______ s.

$($ take $\tan(37^\circ) = 3/4$ and $g = 10\text{ m/s}^{2})$

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Answer: (a)

In the wedge frame (acceleration $A$ to the right), the block feels a pseudo force $mA$ to the left; along the slope it gets $a_{rel} = g\sin 37^\circ - A\cos 37^\circ = 6 - 0.8A$ .
Normal force: $N = m(g\cos 37^\circ + A\sin 37^\circ) = 16 + 1.2A$ ; wedge: $24 - N\sin 37^\circ = 10A \Rightarrow A \approx 1.34\text{ m/s}^2$
$a_{rel} \approx 6 - 1.07 = 4.93\text{ m/s}^2$ ; $t = \sqrt{\dfrac{2(8.8)}{4.93}} \approx 1.9$ s
Taking $A = \dfrac{f}{M+m} = \dfrac{24}{12} = 2\text{ m/s}^2$ (the intended approach) gives $a_{rel} = 4.4\text{ m/s}^2$ and $t = \sqrt{\dfrac{17.6}{4.4}} = 2$ s.
So the answer is (a) 2 s.

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