physics
posted by Kasie on .
A 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table. The bullet passes through block 1 (mass 1.16 kg) and embeds itself in block 2 (mass 1.85 kg). The blocks end up with speeds v1 = 0.560 m/s and v2 = 1.36 m/s (Figure (2)). Neglecting the material removed from block 1 by the bullet, find the speed of the bullet as it (a) enters and (b) leaves block 1.

m =0.0035 kg, m1 =1.16 kg,
m2 = 1.85 kg,
v1 =0.560 m/s, v2 = 1.36 m/s,
u1 =?, u2 =?
When the bullet embeds itself into
block #2, we have a completely inelastic collision. Momentum is conserved and the two objects move together as one. The conservation equation yields:
m•u2 = (m+m2)•v2,
u2 = (m + m2)•v2/m =
= (0.0035+1.85) •1.36/0.0035 =720 m/s,
For the first collision:
m•u1= m•u2 +m1•v1,
u1= (m•u2 +m1•v1)/m =
=(0.0035•720+1.16•0.560)/0.0035 =
= 905.6 m/s, 
Thank You!