Question

In the arrangement shown in the figure, a wedge of mass $m^{3} = 3.45 kg$ is placed on a smooth horizontal surface. A small and light pulley is connected on its top edge as shown. A light flexible thread passes over the pulley. Two blocks having mass $m_{1} = 1.3 kg$ and $m_{2} = 1.5 kg$ are connected at the ends of the thread. $m_{1}$ is on smooth horizontal surface and $m_{2}$ rests on inclined surface of the wedge. Base length of the wedge is $2 m$ and inclination is $370 . m_{2}$ is initially near the top edge of the wedge. If the whole system is released from rest. Calculate:

(i) velocity of wedge when $m_{2}$ reaches its bottom
(ii) velocity of $m_{2}$ at that instant and tension in the thread during motion of $m_{2}$ . All the surfaces are smooth $[g = 10 m s^{- 2}]$

Accepted Answer

Let acceleration of $m$ be a (rightwards) and that of wedge be $b$ (leftwards).

Acceleration of $m_{2}$ (relative to wedge) becomes $(a + b)$ down the plane. Therefore, resultant acceleration of $m_{2}$ is vector sum of the two acceleration.

$\Rightarrow$ Hence components of this resultant acceleration are

$\Rightarrow$ $[(a + b) \cos 37 ° - b] = \frac{(y a - b)}{5}$ horizontal rightwards

$\Rightarrow$ $(a + b) \sin 37 ° = \frac{3}{5} (a + b)$ vertically downwards.

Let us make free body diagrams

For horizontal forces on $m_{1}$ , $T = m_{1} a \dots (1)$

For horizontal forces on wedge, $T - T \cos 37 ° + N_{2} \sin 37 ° = m_{3} b \dots (2)$

For horizontal forces on $m_{2}$ ,

$N_{2} \sin 37 ° - T \cos 37 ° = m_{2} (0.8 a - 0.2 b) \dots (3)$

For vertical forces on $m_{2}$ ,

$m_{2} g - N_{2} \cos 37 ° - T \sin 37 ° = m_{2} (0.8 a - 0.2 b) \dots (4)$

From above equations, $a = 3 {m s}^{- 2}, b = 2 {m s}^{- 2}$ . Since, base angle and base length of wedge are $37 °$ and $2 m$ , therefore, height of its vertical focus is $2 \tan 37 ° = 1.5 m$ .

Now considering vertical motion of $m_{2}$ five top to bottom of the wedge, $μ = 0$

acceleration $= (0.6 a = 0.6 b) = 3 {m s}^{- 2}$ and displacement $= 1.50 m$

Using $S = u t + \frac{1}{2} a t^{2}$

$t = 1 s$

At this instant, horizontal component of velocity $m_{2}$ is

$V_{2 x} = (0.8 a - 0.2 b) t = 2 {m s}^{- 1}$ and vertical component,

$V_{2 y} = (0.6 a + 0.6 b) t = 3 {m s}^{- 1}$

Vertical of $m_{2}$ is

$V_{2} = \sqrt{V_{2 x}^{2} + V_{2 y}^{2}} = \sqrt{13} {m s}^{- 1}$

Velocity of wedge at this instant $= b t = 2 {m s}^{- 1}$

Important Questions on Laws of Motion

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