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
plz help the teacher gave us homework without explaining how to do it
Answers
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
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 = ?.
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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