Energy and Momentum Problem?

Question

Can someone help me with this problem...Thanks
A 1.6 kg block is attached to the end of a 2.0 m string to form a pendulum. The pendulumis released from rest when the string is horizontal. At the lowest point of its swing when it is movinghorizontally, the block is hit by a 10 g bullet moving horizontally in the opposite direction. The bulletremains in the block and causes the block to come to rest at the low point of its swing. What was themagnitude of the bullet’s velocity just before hitting the block?

Answer 1

Using conservation of energy we can find the velocity of the block when it reaches the low poing:
mgh = (1/2)m v^2 Where h is the radius of the swing (2.0)
Rearranged:
v_block = sqrt(2gh) = sqrt(2*9.81m/s^2 * 2m) = 6.26418 m/s

The momentum equation is:
(m_block)(v_block) + (m_bullet)(v_bullet) = 0
Therefore
(m_block)(v_block) = -(m_bullet)(v_bullet)

v_bullet = - (m_block)(v_block)/(m_bullet)
=(1.6kg)(6.26418m/s)/(.010kg)
= -1002.2688 m/s

Where the negative sign indicates the direction of motion.