Ballistic Pendulum Puzzle

Recall that kinetic energy is given by $\frac{1}{2}\times m\times v^{2}$ . Therefore, the kinetic energy of the block-bullet unit at the instant after the collision is given by $\frac{1}{2}\times (m_{1}+m_{2})\times v_{2}^{2}$ .

Also recall that gravitational potential energy is given by $m\times g\times h$ . The block-bullet unit swings up and comes to rest for an instant at height $y$ , at which point its GPE is thus $(m_{1}+m_{2})\times g\times y$ .

Energy conservation gives KE=GPE, yielding $\frac{1}{2}\times (m_{1}+m_{2})\times v_{2}^{2}= (m_{1}+m_{2})\times g\times y$ . Simplifying this yields $v_{2}=\sqrt{2gy}$ .

Now consider the conservation of momentum: $m_{b}\times v_{1}=(m_{b}+m_{w})\times v_{2}$ . We can substitute the expression for $v_{2}$ which we just found into this momentum equation, yielding $v_{1}=\frac{m_{b}+m_{w}}{m_{b}}\times \sqrt{2gy}$ .

Now it is simply the case of inputting the given values of $m_{b}, m_{w}$ and $y$ to find the value of the target variable, $v_{1}$ :

$\frac{0.01+4.00}{0.01}\times \sqrt{2\times 9.8\times 0.04}=355.06...ms^{-1}$

So, to 3 significant figures and without units, the solution is $\boxed{355}$ .