Q. Mass of 3 kg descending vertically downward supports a mass of 2 kg by means the end of 5 s, the string breaks. How much higher the 2 kg mass will go further?

A) 4.9 m
B) 9.8 m
C) 19.6 m
D) 2.45 m
Explanation

(a) Acceleration of combined system,
$a = \frac{m_{1} - m_{2}}{m_{1} + m_{2}}.g$
$= \frac{3 - 2}{3 + 2} \times 9.8 = 1.96\ ms^{- 2}$
Vertically upward velocity of 2 kg mass at the time breaking of string,
$v = at = 5 \times 1.96 = 9.8\ ms^{- 2}$
After breaking of string, mass $m_{2}$ moves under gravity and go further higher through a height $h$, where final velocity is zero. Hence
$(0)^{2} - (9.8)^{2} = 2 \times ( - 9.8) \times h$ or $h = 4.9\ $m

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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