Question

A bullet of mass $mkg$ and horizontal speed $um{s}^{-1}$ strikes a block of wood of mass $Mkg$ and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Answer 1

The initial total energy (kinetic + potential) = the final total energy (kinetic + potential)

$K{E}_0+P{E}_0=K{E}_f+P{E}_f$

Right after the bullet hits the block, the initial height is 0, so $P{E}_0=0$. When the bullet/block reaches its maximum height, it is at rest, so $K{E}_f=0$

$K{E}_0=P{E}_f$

$\frac{1}{2}\left(m+M\right)\times {v_f}^2=\left(m+M\right)\times gh$

Equation to solve for "$v_f$, the velocity of the bullet/block after the collision:

$v_f=\sqrt{2gh}$

For an inelastic collision (where the objects become stuck together), the initial momentum = final momentum.

$p_0=p_f$

$m_{bullet}\times v_{0bullet}+M_{block}\times v_{0block}=\left(m_{bullet}+M_{block}\right)\times v_f$

Since the block is not moving before the collision,$v_{0block}=0$

$m_{bullet}\times v_{0,bullet}=\left(m+M\right)\times v_f$

Rearranging to solve for in initial velocity of the bullet, "$v_0$":

$v_0=\frac{\left(m+M\right)\times v_f}{m}$

Substituting in the expression for "$v_f$" solved for:

$v_0=\frac{\left(m+M\right)\times \sqrt{2gh}}{m}$