Question

A 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

Answer 1

Answer: (A)

Acceleration of combined system,
$a=\frac{m_{1}-m_{2}}{m_{1}+m_{2}} \cdot g=\frac{3-2}{3+2} \times 9.8=1.96\, ms ^{-2}$
Vertically upward velocity of $2 \,kg$ mass at the time of breaking of string, $v=a t=5 \times 1.96=9.8 \,ms ^{-2}$.
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 $h =4.9 \,m$