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Laws of Motion

JEE Main level, single correct, medium

4 of 210

Question 4 of 210

A wedge $Y$ with mass of $10$ kg and all frictionless surfaces and the inclined surface making $37°$ 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 $\tan(37°) = 3/4$ and $g = 10$ m/s $^2$ )

Show answer and solution

Answer: (A) $2$

Horizontal acceleration of the wedge (assuming the block and wedge move together horizontally):

A = \dfrac{f}{M + m} = \dfrac{24}{10 + 2} = 2\text{ m/s}^2

Working in the non-inertial frame of the wedge, the block experiences a pseudo force

mA

directed opposite to the wedge's acceleration. The components along the incline are:
Component of gravity down the incline:

g\sin 37^\circ

Component of pseudo force up the incline:

A\cos 37^\circ

Net relative acceleration of the block down the incline:

a_{\text{rel}} = g\sin 37^\circ - A\cos 37^\circ

Using

\sin 37^\circ = 0.6

and

\cos 37^\circ = 0.8

a_{\text{rel}} = 10(0.6) - 2(0.8) = 6 - 1.6 = 4.4\text{ m/s}^2

Time taken to slide a distance of

8.8\text{ m}

along the incline, starting from rest:

s = \dfrac{1}{2} a_{\text{rel}}\, t^2

8.8 = \dfrac{1}{2}(4.4)\,t^2

8.8 = 2.2\,t^2

t^2 = 4 \Rightarrow t = 2\text{ s}

Hence, the required time is

2\text{ s}

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