Asked by flirtyme

In an experiment to find the speed of 2.4g bullet, it was fired into a 650g block of wood at rest on a friction free surface. If the block(and bullet) moved off with an initial speed of 96cm s-1, what was the speed of the bullet?
plz help the teacher gave us homework without explaining how to do it
Answered by Damon
in kg, meters, seconds:
bullet mass = 0.0024 kg
0.0024 v = (0.650 + 0.0024)(0.96)
v will be in METERS/second
Answered by Henry
Given:
M1 = 0.0024kg, V1 = ?.
M2 = 0.650kg, V2 = 0.
V3 = 0.96 m/s = Velocity of M1 and M2 after the c024V1ollision.

Momentum before = Momentum after:
M1*V1 + M2*V2 = M1*V3 + M2*V3,
0.0024*V1 + 0.65*0 = 0.0024*0.96 + 0.65*0.96,
V1 = ?.

Answered by Henry
V3 = 0.96 m/s = Velocity of M1 and M2 after the collision.
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