Asked by Joseph
A bullet of mass 50g is fired with a velocity of (500,0)m/s in to a sack of sand of mass 20kg which is swinging from rope. At the moment the bullet hits, the sack has a velocity of (0,3)m/s. Workout the velocity of bullet and sack just after the bullet hits the sack.
Answers
Answered by
oobleck
you don't say in which direction the sack is swinging. It makes a difference.
Answered by
Joseph
it was swinging vertically upwards
Answered by
henry2,
Given: M1 = 0.05kg, V1= 500m/s[0o].
M2 = 20 kg, V2 = 3m/s[90o]. = 3i m/s.
V3 = velocity of M1 and M2 after collision.
Momentum before = Momentum after
M1*V1+M2*V2 = M1*V3+M2*V3
0.05*500+20*3i = 0.05V3+20*V3
25 + 60i = 20.05V3
V3 = 1.25 + 2.99i = 3.24m/s[67.3o].
M2 = 20 kg, V2 = 3m/s[90o]. = 3i m/s.
V3 = velocity of M1 and M2 after collision.
Momentum before = Momentum after
M1*V1+M2*V2 = M1*V3+M2*V3
0.05*500+20*3i = 0.05V3+20*V3
25 + 60i = 20.05V3
V3 = 1.25 + 2.99i = 3.24m/s[67.3o].
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