Question

A wedge of mass $M$ is kept at rest on a horizontal surface, and a block of mass $m$ is kept on the wedge at the height $h$ from the surface. The angle of the wedge is $\alpha$, and there is no friction neither between the block and wedge nor between the surface and the wedge. The system is released into a free motion. If the time taken by the block to reach the horizontal surface is $\sqrt{\frac{nh}{g}}$, then the value of $n$ is _____.

Answer 1

3

Answer 2

Let $a_m$ be the acceleration of the block along the incline relative to the wedge, and let $A$ be the magnitude of the wedge's acceleration to the left.

The forces acting on the block along the incline are:

1. Component of gravity: $mg \sin \alpha$
2. Component of fictitious force due to wedge's acceleration $A$ (acting in the direction of wedge's acceleration, i.e., to the left, which is also along the incline direction): $mA \cos \alpha$. The net force along the incline is $m a_m = mg \sin \alpha + mA \cos \alpha$.

The forces acting on the wedge are:

1. Horizontal component of the normal force from the block: $N \sin \alpha$ (acting to the right). The equation of motion for the wedge (accelerating to the left with magnitude $A$) is $MA = N \sin \alpha$.

The normal force $N$ exerted by the wedge on the block is perpendicular to the incline. In the frame of the wedge, the block is not accelerating perpendicular to the incline. The forces perpendicular to the incline are:

1. Component of gravity: $-mg \cos \alpha$
2. Normal force $N$ (acting outwards, perpendicular to the incline). So, $N - mg \cos \alpha = 0$, which gives $N = mg \cos \alpha$.

Substitute $N$ into the wedge's equation of motion: $MA = (mg \cos \alpha) \sin \alpha = mg \sin \alpha \cos \alpha$. $A = \frac{m}{M} g \sin \alpha \cos \alpha$.

Now substitute $A$ into the block's equation of motion along the incline: $m a_m = mg \sin \alpha + m \left(\frac{m}{M} g \sin \alpha \cos \alpha\right) \cos \alpha$ $a_m = g \sin \alpha + \frac{m^2}{M} g \sin \alpha \cos^2 \alpha$ $a_m = g \sin \alpha \left(1 + \frac{m^2}{M} \cos^2 \alpha\right)$.

The block slides down a distance $s = h / \sin \alpha$ along the incline. Since the acceleration $a_m$ is constant, we can use the kinematic equation $s = \frac{1}{2} a_m t^2$. $t^2 = \frac{2s}{a_m} = \frac{2(h/\sin \alpha)}{g \sin \alpha \left(1 + \frac{m^2}{M} \cos^2 \alpha\right)}$ $t^2 = \frac{2h}{g \sin^2 \alpha \left(1 + \frac{m^2}{M} \cos^2 \alpha\right)}$ $t^2 = \frac{2h}{g \left(\sin^2 \alpha + \frac{m^2}{M} \sin^2 \alpha \cos^2 \alpha\right)}$.

Given $m = 2$ kg, $M = 6$ kg, $\alpha = 45^\circ$. $\sin \alpha = \sin 45^\circ = \frac{1}{\sqrt{2}}$, $\sin^2 \alpha = \frac{1}{2}$. $\cos \alpha = \cos 45^\circ = \frac{1}{\sqrt{2}}$, $\cos^2 \alpha = \frac{1}{2}$. $\frac{m^2}{M} = \frac{2^2}{6} = \frac{4}{6} = \frac{2}{3}$.

$t^2 = \frac{2h}{g \left(\frac{1}{2} + \frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{2}\right)}$ $t^2 = \frac{2h}{g \left(\frac{1}{2} + \frac{2}{12}\right)} = \frac{2h}{g \left(\frac{1}{2} + \frac{1}{6}\right)}$ $t^2 = \frac{2h}{g \left(\frac{3+1}{6}\right)} = \frac{2h}{g \left(\frac{4}{6}\right)} = \frac{2h}{g \left(\frac{2}{3}\right)}$ $t^2 = \frac{2h \cdot 3}{2g} = \frac{3h}{g}$.

The problem states $t = \sqrt{\frac{nh}{g}}$. Comparing $t^2 = \frac{3h}{g}$ with $t^2 = \frac{nh}{g}$, we get $n=3$. The value of $n$ is 3.