Question

A bag of sand mass M is suspended by a string. A bullet of mass $m$ is fired at it with velocity v and gets embedded into it. The loss of kinetic energy in this process is:

• A. $\frac{1}{2}m{v}^2$
• B. $\frac{1}{2}m{v}^2\times \frac{1}{M+m}$
• C. $\frac{1}{2}m{v}^2\times \frac{M}{m}$
• D. $\frac{1}{2}m{v}^2\left(\frac{M}{M+m}\right)$

Answer 1

The correct option is D $\frac{1}{2}m{v}^2\left(\frac{M}{M+m}\right)$

Initially, only the bullet is moving, and so the total momentum of the bullet+bag system is mv. Once the bullet is embedded in the bag, the bullet and bag are moving together, and therefore have the same velocity (call it u). So, by conservation of momentum, we have

$(M+m)u=mv$

$u=\frac{mv}{M+m}$

We now have what we need to compare the initial and final kinetic energies. Probably the most useful way to do this is as a ratio: the final kinetic energy, as a fraction of the initial kinetic energy, is

$\frac{m+M\frac{u^2}{2}}{\frac{m{v}^2}{2}}=\frac{M+m}{m}{\left(\frac{u}{v}\right)}^2=\frac{m+M}{m}{\left(\frac{m}{m+M}\right)}^2=\frac{m}{m+M}$

We could also see this by noting that the kinetic energy of a single moving mass is equal to $\frac{p^2}{2m}$, and that p isn’t changing during the collision, so the kinetic energy will vary inversely with (uniformly moving) mass.

$K{E}_{loss}=K{E}_{ini}(1-\frac{m}{m+M})=\frac{1}{2}m{v}^2\left(\frac{m}{m+M}\right)$