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! :(

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.