Question

A cannon of mass $10\times {10}^{3}kg$ is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a $2.1\times {10}^{3}kg$ shell horizontally with an initial velocity of $550m/s$. Suppose the cannon is then unbolted from the earth and no external force hinder its recoil. What would be the velocity (in $m/s$) of a shell fired horizontally by the loose cannon? (Hint: In both cases assume that the burning gunpowder imparts the same kinetic energy to the system.)

Answer 1

Kinetic energy in the first case :$\frac{1}{2}m{v}^2$

$=\frac{1}{2}(2.1\times {10}^3)(550{)}^2=3.17\times {10}^{8}J$

In the second case,we have to conserve both momentum and kinetic energy between both cannon and shell:

$p_{before}=0=p_{after}=m_{shell}{v}_{shell}-m_{cannon}{v}_{cannon}$

$m_{cannon}{v}_{cannon}=m_{shell}{v}_{shell}$

$v_{cannon}=\frac{m_{shell}{v}_{shell}}{m_{cannon}}$

$k=\frac{1}{2}{m}_{shell}{v}_{shell}^2+\frac{1}{2}{m}_{cannon}{v}_{cannon}^2$

$k=\frac{1}{2}{m}_{shell}{v}_{shell}^2+\frac{1}{2}{m}_{cannon}(\frac{m_{shell}{v}_{shell}}{m_{cannon}}{)}^2$

$k=\frac{1}{2}{m}_{shell}{v}_{shell}^2+\frac{1}{2}{m}_{shell}{v}_{shell}^2(\frac{m_{shell}}{m_{cannon}}{)}^2$

$=\frac{1}{2}{m}_{shell}{v}_{shell}^2(1+(\frac{m_{shell}}{m_{cannon}}$))

$v_{shell}^2=\frac{2k}{m_{shell}(1+(\frac{m_{shell}}{m_{cannon}}\left)\right)}$

$v_{shell}=\sqrt{\frac{2\times 3.17\times {10}^8}{(2.1\times {10}^3)(1+\frac{2.1}{10})}}=\sqrt{249508}\cong 499.5=500m/s$