A mass of $3 k g$ descending vertically downward supports a mass of $2 k g$ by means the end of $5 s$ , the string breaks. How much higher the $2 k g$ mass will go further?

4.9 m

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9.8 m

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19.6 m

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2.45 m

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Solution

The correct option is A $4.9 m$
Acceleration of the system before the string breaks
$a = \frac{net pulling force}{total mass} = \frac{3 g - 2 g}{(3 + 2)}$
$= \frac{g}{5}$
after $5 s$ , velocity of the system,
$V = a t = \frac{g}{5} \times 5 = g$ m/s
required height $h = \frac{v^{2}}{2 g} = \frac{g \times g}{2 g} = \frac{g}{2} = \frac{9.8}{2} = 4.9$ m

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