Posted by **Kasie** on Wednesday, May 9, 2012 at 3:46pm.

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.

- physics -
**Elena**, Wednesday, May 9, 2012 at 4:43pm
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,

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**Skywalker**, Saturday, October 18, 2014 at 10:09pm
Thank You!

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