A 6-g bullet is fired into a 1.2-kg block that is resting on a table. After impact, the block with the embedded bullet slides 7 cm across the table before coming to rest again. The coefficient of kinetic friction between the block and the surface of the table is 0.68. What was the initial speed of the bullet? THANK YOU :) please, explain!!!

Question

anonymous · Answer

Not a full algebraic solution but here is my attempt..

Make sure you use conservation of momentum here, i.e. 
$$m _{Bullet}v _{Bullet} = m _{tot}v _{tot}$$
You will need to then substitute the v_total for an expression, given how far the block travels (use F=ma, and v^2 = u^2 +2as), here v = 0 but u is your$$v_{tot}$$
Sub it all in and you should be able to solve for the bullet velocity (my answer comes out as 29.1 m/s, which seems slow (108km/h)).