Name: Conservation of Linear Momentum
Uploaded: 2019-07-19
Duration: 3 min 5 s
Description: In their study of natural phenomena over many centuries, scientists have identified certain fundamental laws. One such law is the law of conservation of mome...

Question

In a simple Atwood machine, two unequal masses $m_{1}$ and $m_{2}$ are connected by a string going over a clamped light smooth pulley. In a typical arrangement as shown in the figure, $m_{1} = 300 g$ and $m_{2} = 600 g$ . The system released from rest and then the larger mass is stopped for a moment $2.0 s$ after the system is set into motion. Find the time elapsed before the string is tight again.

Accepted Answer

Given, the masses of the blocks are $m_{1} = 300 g = 0.3 kg$ and $m_{2} = 600 g$ .

Let $T$ be the tension in the string and $a$ be the common acceleration of the blocks.

The free-body diagrams of both the masses will be,

According to Newton's second law, equation for mass $m_{1}$ will be,

$T - m_{1} g = m_{1} a \dots (1)$

According to Newton's second law, equation for mass $m_{2}$ will be,
$m_{2} g - T = m_{2} a \dots (2)$

Adding equations $(1)$ and $(2)$ , we get,

$g (m_{2} - m_{1}) = a (m_{1} + m_{2})$

$\Rightarrow a = \frac{g (m_{2} - m_{1})}{m_{2} + m_{1}} = \frac{9.8 \times 0.3}{0.9} = 3.266 m s^{- 2}$ .

At $t = 2 s$ , velocity of $m_{1}$ will be,

using the first equation of motion,

$v = u + a t$

$\Rightarrow v = 3.266 \times 2 = 6.532 m s^{- 1}$ .

After $2$ seconds, the larger mass is stopped for a moment so the string will become slack afterwards, but the smaller mass will continue to move upwards until it's velocity becomes zero.

Again using the first equation of motion,

here, $a = g = - 9.8 m s^{- 2}$ .

$v = u + a t$

$\Rightarrow 0 = 6.532 - 9.8 t$

$\Rightarrow t = \frac{6.532}{9.8} = 0.666 s$ .

Since both are free-falling bodies, the distance travelled by of them will be same, hence, slackness will become zero at $t = 0.666 s$ .

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