An artillery piece of mass M_1 fires a shell | vedclass

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Question

(B) According to the law of conservation of linear momentum,the magnitude of the momentum of the artillery piece $(p_1)$ and the shell $(p_2)$ must be

Vedclass · Accepted Answer

$M_2 / M_1$

Vedclass · Answer

(B) According to the law of conservation of linear momentum,the magnitude of the momentum of the artillery piece $(p_1)$ and the shell $(p_2)$ must be equal immediately after firing,as the initial momentum of the system was zero.$|p_1| = |p_2| = p$Kinetic energy $(KE)$ is related to momentum $(p)$ and mass $(m)$ by the formula $KE = \frac{p^2}{2m}$.Since the momentum $p$ is the same for both,$KE \propto \frac{1}{m}$.Therefore,the ratio of the kinetic energy of the artillery $(KE_1)$ to that of the shell $(KE_2)$ is:$\frac{KE_1}{KE_2} = \frac{p^2 / 2M_1}{p^2 / 2M_2} = \frac{M_2}{M_1}$.

An artillery piece of mass $M_1$ fires a shell of mass $M_2$ horizontally. Instantaneously after the firing,the ratio of kinetic energy of the artillery and that of the shell is:

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