Two identical trolley, A and B, collide.
The two trolley each weight 0.5kg
Before collision,
Trolley A move right with speed of 0.5m/s
and Trolley B was at rest
After collision,
The two trolley "stick" together
and move rightwards.
1) What is the speed of trolley A and B after impact?
2) If time of impact is 0.3 seconds, what is the average force acting on the two trolleys?
3) Will the average force acting on trolley A greater than that of trolley B if troley B's mass is higher than trolley A's mass? Explain.
Physics - momentum
2012-03-04 11:54 pm
回答 (2)
2012-03-05 10:45 pm
✔ 最佳答案
1. Since total momentum of a system is always conserved during a collusion. Taking right direction be positive.
Σ(mu) = Σ(mv)
0.5(0.5) = (0.5+0.5)v
v = 0.25 m/s.
As they stick together, they move with a common velocity of 0.25 m/s to the right. So the speed is 0.25 m/s.
2.
By Newton's second law of motion,
ΣF = ma
ΣF = m(dv/dt)
ΣF = dp/dt where p is the momentum
Consider trolley A:
We have, Favg = 0.5(0.25 - 0.5) / 0.3 = -0.417 N
The average force acting on A is 0.417 N to the left.
By Newton's third law of motion, the average force acting on B is 0.417 N to the right.
3.
No. By Newton's third law of motion, the force on A is by B. The force on B is by A. They are collinear and equal in magnitude, but opposite in direction.
2012-03-05 14:48:14 補充：
The second Q takes 3 s.f.
參考: My physics knowledge
2012-03-05 12:23 am
1. By conservation of momentum,
m1u1+m2u2=m1v1+m2v2
(0.5)(0.5)+(0.5)(0)=(0.5)(v)+(0.5)(v)
v=0.25m/s
2.F=(mv-mu)/t
=((0.5)(0.25)-(0.5)(0))/(0.3)
=0.417N
3. No. There are action and reaction pairs.
m1u1+m2u2=m1v1+m2v2
(0.5)(0.5)+(0.5)(0)=(0.5)(v)+(0.5)(v)
v=0.25m/s
2.F=(mv-mu)/t
=((0.5)(0.25)-(0.5)(0))/(0.3)
=0.417N
3. No. There are action and reaction pairs.
參考: 自己
收錄日期: 2021-04-20 16:31:02
原文連結 [永久失效]:
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