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Physics Question on work, energy and power

A shell of mass 200 gm is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is

A

40 ms−1^{-1}

B

120 ms−1^{-1}

C

100 ms−1^{-1}

D

80 ms−1^{-1}

Answer

100 ms−1^{-1}

Explanation

Solution

mv = Mv' ⇒    v′=(mM)v\Rightarrow \, \, \, \, v' =\bigg(\frac{m}{M}\bigg)v
Total K.E. of the bullet and gun
=12mv2+12Mv′2=\frac{1}{2}mv^2+\frac{1}{2}Mv'{^2}
Total K.E =12mv2+12M⋅m2M2v2\frac{1}{2}mv^2+\frac{1}{2}M\cdot \frac{m^2}{M^2}v^2
Total K.E =\frac{1}{2}mv^2 \bigg \\{1+\frac{m}{M}\bigg \\}
\, \, \, \, \, = \bigg \\{ \frac{1}{2} \times \, 0.2 \bigg \\} \bigg\\{ 1+\frac{0.2}{4}\bigg\\} v^2=1.05 \times 1000\, J
⇒    v2=4× 1.05 ×10000.1 × 4.2=1002\Rightarrow \, \, \, \, v^2=\frac{4 \times \, 1.05 \, \times 1000}{0.1 \, \times \, 4.2}=100^2;
∴   v=100 ms−1\therefore \, \, \, v=100 \,ms^{-1}