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pleasehelpmetoo · May 13, 2009, 03:09 AM

A 10g bullet moving at 400 m/s penetrates a 3.okg block of wood resting on a frictionless horizontal surface. If the bullet slows down to 300m/s,after emerging from the block, find the speed of the block immediately after the bullet emerges.

does it involve use of eqn: v=u +at or Force =mass xacceleration?
initial velocity 400
final velocity 300
where does the mass of bullet and wooden block come into the equation?
Help PLEASE! :(

Unknown008 · May 13, 2009, 06:43 AM

No, this time you have to use the law of conservation of momentum, given by the formula:

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

$m_1$ is the mass of the bullet

$u_1$ is the initial speed of the bullet

$m_2$ is the mass of the block

$u_2$ is the initial speed of the block

$v_1$ is the final speed of the bullet

$v_2$ is the final speed of the block.

Solve for $v_2$ to have the final speed of the block.