Question

A 20 gram bullet moving horizontally with a velocity of 550 m/s gets imbedded in a 12 kg block hanging by means of a light cord. To what vertical height will the block rise?

Answer 1

Let's apply the Law of Conservation of Momentum and find the velocity of the combination of block and bullet when the bullet gets embedded in it:

$$m_{bullet}v_{bullet}+m_{block}v_{block}=(m_{bullet}+m_{block})v,$$

here, $m_{bullet}=0.02\ kg$ is the mass of the bullet, $v_{bullet}=550\ \dfrac{m}{s}$ is the velocity of the bullet before the collision, $m_{block}=12\ kg$ is the mass of the block, $v_{block}=0\ \dfrac{m}{s}$ is the velocity of the block before the collision, $v$ is the velocity of the combination of block and bullet after the collision.

Then, from this equation we can find $v$:

$$v=\dfrac{m_{bullet}v_{bullet}}{(m_{bullet}+m_{block})},$$ $$v=\dfrac{0.02\ kg\cdot 550\ \dfrac{m}{s}}{(0.02\ kg+12\ kg)}=0.91\ \dfrac{m}{s}.$$

We can find the vertical height attained by the block from the Law of Conservation of Energy:

$$KE=PE,$$ $$\dfrac{1}{2}(m_{bullet}+m_{block})v^2=(m_{bullet}+m_{block})gh,$$ $$h=\dfrac{v^2}{2g}=\dfrac{(0.91\ \dfrac{m}{s})^2}{2\cdot 9.8\ \dfrac{m}{s^2}}=0.043\ m.$$

Answer:

$h=0.043\ m.$