A 12-g bullet moving horizontally strikes and remains in a 3.0-kg block initially at rest on the edge of a table. The block, which is initially 80 cm above the floor, strikes the floor a horizontal distance of 120 cm from its initial position. What was the initial speed of the bullet?

Answer 1

Δy = v0t + ½ at2
- 0.8 = (0)t + ½ (-9.8)t2
0.404 s = t
Δx = v0t + ½ at2
1.2 = v0 (0.404) + ½ (0)(0.404)2
2.97 m/s = v0 (after the collision as the block leaves the table)
p0 = p1
(0. 012 kg )v + (3 kg)(0 m/s) = (0.012 kg)(2.97 m/s) + (3 kg)(2.97 m/s)
(0. 012 kg )v = 8.97
745.5 m/s = v