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Question

(A) Let the mass of the incoming object be $m$ and its velocity be $v$. The kinetic energy is $E = \frac{1}{2}mv^2$ and momentum is $P = mv$. Thus,$m

Vedclass · Accepted Answer

First increases then decreases with linear momentum

Vedclass · Answer

(A) Let the mass of the incoming object be $m$ and its velocity be $v$. The kinetic energy is $E = \frac{1}{2}mv^2$ and momentum is $P = mv$. Thus,$m = \frac{P^2}{2E}$.For a head-on elastic collision with a stationary mass $M$,the kinetic energy transferred to $M$ is given by $\Delta K = \frac{4mM}{(m+M)^2} E$.Substituting $m = \frac{P^2}{2E}$ into the expression:$\Delta K = \frac{4(\frac{P^2}{2E})M}{(\frac{P^2}{2E} + M)^2} E = \frac{2P^2 M}{(\frac{P^2 + 2ME}{2E})^2} E = \frac{8ME P^2}{(P^2 + 2ME)^2}$.Let $f(P) = \frac{8ME P^2}{(P^2 + 2ME)^2}$. As $P 	o 0$,$f(P) 	o 0$. As $P 	o \infty$,$f(P) 	o 0$. Since $f(P) > 0$ for $P > 0$,the function must increase to a maximum and then decrease. Therefore,the energy transferred first increases and then decreases with linear momentum $P$.

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