Question

A bullet of m=8.00g is fired into a block of mass M=240g that is initially at rest at the edge of a table of height h=1.00m.The bullet remains in the block,and after the impact the block lands d=1.80 m from the bottom of the table.Determine the initial speed of the bullet.

Answer 1

Let's first find the time that the combination of block and bullet takes to reach the ground:

$$y=\dfrac{1}{2}gt^2,$$ $$t=\sqrt{\dfrac{2y}{g}}=\sqrt{\dfrac{2\times1.0\ m}{9.8\ \dfrac{m}{s^2}}}=0.45\ s.$$

Then, we can find the initial velocity of the combination of block and bullet before it falling down from the table:

$$x=d=v_0t,$$ $$v_0=\dfrac{x}{t}=\dfrac{1.8\ m}{0.45\ s}=4\ \dfrac{m}{s}.$$

Finally, we can find the initial velocity of the bullet from the law of conservation of momentum:

$$m_bv_{i,b}=(m_b+m_{block})v_c,$$ $$v_{i,b}=\dfrac{(m_b+m_{block})v_c}{m_b},$$ $$v_{i,b}=\dfrac{(0.008\ kg+0.24\ kg)\times4\ \dfrac{m}{s}}{0.008\ kg}=124\ \dfrac{m}{s}.$$