Answer
The block rises to a maximum height of $~~7.35~cm$
Work Step by Step
We can use conservation of momentum to find the speed of the block after the bullet passes through:
$p_f = p_i$
$(5000~g)(v)+(10~g)(400~m/s) = (10~g)(1000~m/s)$
$(5000~g)(v) = (10~g)(1000~m/s)-(10~g)(400~m/s)$
$v = \frac{(10~g)(1000~m/s)-(10~g)(400~m/s)}{5000~g}$
$v = 1.2~m/s$
In the next part of the solution, we can let $v_0 = 1.2~m/s$
We can find the block's maximum height:
$v_f^2 = v_0^2+2ay$
$y = \frac{v_f^2 - v_0^2}{2a}$
$y = \frac{0 - (1.2~m/s)^2}{(2)(-9.8~m/s^2)}$
$y = 0.0735~m$
$y = 7.35~cm$
The block rises to a maximum height of $~~7.35~cm$
