A bullet of mass 0.01 kg is moving horizontally with a speed of 100 m/s when it hits a block of mass 2 kg that is at rest on a horizontal surface with a coefficient of friction of 0.4. After the collision the bullet becomes embedded in the block.

Answer 1

initial momentum = 0.01 * 100

momentum immediately after collision = 2.01 Vi

so
2.01 Vi = 0.01 * 100
solve for Vi

Then you have a force of F = -2.01 * g * 0.4 retarding the motion

a = F/m = F/2.01 = -0.4 g

V = Vi -0.4 g t
x = Vi t -(0.4 g/2)t^2