Asked by Yonas
A 50-g bullet with a speed of 10 m/s hits into a sandbag hanging from the ceiling and embeds into it. The sandbag is at rest before the collision and gain a speed of 0.2 m/s afterwards. How much does the sandbag weigh?
Answers
Answered by
scott
momentum is conserved
initial momentum (bullet) , equals final momentum (bullet + sandbag)
50 g * 10 m/s = (S + 50 g) * 0.2 m/s
do you need to differentiate between mass and weight?
initial momentum (bullet) , equals final momentum (bullet + sandbag)
50 g * 10 m/s = (S + 50 g) * 0.2 m/s
do you need to differentiate between mass and weight?
Answered by
Henry
Given:
M1 = 0.05 kg, V1 = 10 m/s.
M2 = ?, V2 = 0.
V3 = Velocity of M1 after collision.
V4 = 0.2 m/s = velocity of M2 after collision.
Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*V4,
0.05*10 + M2*0 = 0.05*0 + M2*0.2,
0.5 = 0.2M2,
M2 = 2.5 kg. = Wt. of sandbag.
M1 = 0.05 kg, V1 = 10 m/s.
M2 = ?, V2 = 0.
V3 = Velocity of M1 after collision.
V4 = 0.2 m/s = velocity of M2 after collision.
Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*V4,
0.05*10 + M2*0 = 0.05*0 + M2*0.2,
0.5 = 0.2M2,
M2 = 2.5 kg. = Wt. of sandbag.
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