A 4.25-g bullet traveling horizontally with a velocity of magnitude 375 m/s is fired into a wooden block with mass 1.12 kg, initially at rest on a level frictionless surface. The bullet passes through the block and emerges with its speed reduced to 114 m/s. Part A How fast is the block moving just after the bullet emerges from it

0.9904 m/s

Explanation:

To solve this problem we need to use the conservation of momentum:

m1v1 + m2v2 = m1'v1' + m2'v2'

Using this equation, we have:

m_bullet*v_bullet = m_bullet_after*v_bullet_after + m_block*v_block

0.00425 * 375 = 0.00425 * 114 + 1.12 * v_block

1.12 * v_block = 1.5938 - 0.4845

1.12 * v_block = 1.1093

v_block = 1.1093 / 1.12

v_block = 0.9904 m/s

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.