Question

1) A spring ball of mass 0.5 kg is dropped
from some height. On falling freely for
10 s, it explodes into two fragments of
mass ratio 1:2. The lighter fragment
continues to travel downwards with
speed of 60 m/s. Calculate the kinetic
energy supplied during explosion​

Answer 1

Mass of the spring ball,

• $\sf{M=0.5\ kg}$

Mass of the two fragments will be,

• $\sf{m_1=\dfrac{0.5}{3}\ kg}$

• $\sf{m_2=\dfrac{0.5\times2}{3}\ kg}$

As the spring ball is dropped, by first equation of motion, the speed of the spring ball after 10 seconds of dropping,

$\sf{\longrightarrow V=u+gt}$

$\sf{\longrightarrow V=0+9.8\times10}$

$\sf{\longrightarrow V=98\ m\,s^{-1}}$

The speed of the lighter fragment,

• $\sf{v_1=60\ m\,s^{-1}}$

Let $\sf{v_2}$ be the velocity of the heavier fragment.

Since the linear momentum of the system is conserved,

$\sf{\longrightarrow MV=m_1v_1+m_2v_2}$

$\sf{\longrightarrow 0.5\times98=\dfrac{0.5}{3}\times60+\dfrac{0.5\times2}{3}\,v_2}$

$\sf{\longrightarrow 98=20+\dfrac{2v_2}{3}}$

$\sf{\longrightarrow v_2=117\ m\,s^{-1}}$

Kinetic energy of the spring ball,

$\sf{\longrightarrow K_b=\dfrac{1}{2}\,MV^2}$

$\sf{\longrightarrow K_b=\dfrac{1}{2}\times0.5\times98^2}$

$\sf{\longrightarrow K_b=2401\ J}$

Kinetic energy of the lighter fragment,

$\sf{\longrightarrow K_1=\dfrac{1}{2}\,m_1(v_1)^2}$

$\sf{\longrightarrow K_1=\dfrac{1}{2}\times\dfrac{0.5}{3}\times60^2}$

$\sf{\longrightarrow K_1=300\ J}$

Kinetic energy of the heavier fragment,

$\sf{\longrightarrow K_2=\dfrac{1}{2}\,m_2(v_2)^2}$

$\sf{\longrightarrow K_2=\dfrac{1}{2}\times\dfrac{0.5\times2}{3}\times117^2}$

$\sf{\longrightarrow K_2=2281.5\ J}$

The kinetic energy supplied during explosion is equal to the change in kinetic energy.

Then,

$\sf{\longrightarrow \Delta K=(K_1+K_2)-K_b}$

$\sf{\longrightarrow \Delta K=300\ J+2281.5\ J-2401\ J}$

$\sf{\longrightarrow\underline{\underline{\Delta K=180.5\ J}}}$