First slide

Connected bodies

Question

A mass of 3 kg descending vertically downwards supports a mass of 2 kg by means of a light string
passing over a pulley. At the end of 5 s, the string breaks . How much high from now the 2 kg mass
will go? (Take, 9 = 9.8 $m s^{- 2}$ )

Moderate

A

4.9 m
B

9.8 m
C

19.6 m
D

2.45 m

Solution

Acceleration of system before breaking the string,
$a = \frac{Net pulling force}{Total mass} = \frac{3 g - 2 g}{5} = \frac{g}{5}$
After 5 s velocity of system, $v = a t = \frac{g}{5} \times 5 = g {ms}^{- 1}$
Now, $h = \frac{v^{2}}{2 g} = \frac{g^{2}}{2 g} = \frac{g}{2} = 49 m$

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