1. A can is stationary on a frictionless table. A small stone travels towards the right and hits the can at 2 m/s. The mass of the stone and that of the can are 0.5 kg and 0.4 kg respectively.
a) The velocity of the stone decreases from 2 m/s to 1 m/s after hitting the can for 0.05 s. What is the force acting on the stone during the collision?
b) What is the force acting on the can during the collision?
c) What is velocity of the can after the collision?
2. On 12 May 2005, a traffic accident happened on Kwai Chung Road. A box truck P of 5.5*10^3 kg was broken down and parked in the fast lane. The driver of another box truck Q did not notice truck P. He ran into truck P and both trucks moved forwards for 30 m after the collision.
Assume the mass of truck Q was 3000 kg. The friction acting on each truck was 3000 N during the collision and the friction acting on the trucks was 6000 N when they moved together.
a) What was the speed of the trucks after the collision?
b) If the time of the collision was 0.05 s, what was the force acting on truck P by truck Q?
c) What was the force acting on truck Q by truck P?
d) Find the speed of truck Q before the collision.
Physics 溫習題目 3
2015-06-17 2:51 am
回答 (2)
2015-06-17 5:32 am
✔ 最佳答案
1.a)
F = (mv-mu)/t = (0.5×1 - 0.5×2)/0.05 = -10N
∴ The force acting on the stone is 10N backward.
b)
By Newton's 3rd law,
force acting on the stone = -force acting on the can
∴ The force acting on the can during the collision is 10N to the front.
c)
By the law of conservation of momentum,
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.5×2 + 0 = 0.5×1 + 0.4v
v = 1.25ms⁻¹
∴ The velocity of the can after the collision is 1.25ms⁻¹ to the front.
2.
a)
Assume the two cars as one body. After the collision, by Newton's 2nd law,
F = ma
6000 = (5.5×10³+3000) a
a = 6000/8500 = 12/17 ms⁻²
a = 12/17 ms⁻², v = 0, s = 30m
By v² = u² + 2as
0 = u² + 2(12/17)(30)
u = 6.51 ms⁻¹
∴ The speed of the trucks after the collision was 6.51 ms⁻¹ to the front.
b)
By Newton's 2nd law,
F - f = ma = (mv-mu)/0.05
F - 3000 = (5.5×10³×6.51 - 0)/0.05
F = 719000N
∴ Force acting on truck P by truck Q is 719000N to the front.
c)
By Newton's 3rd law,
force acting on truck Q by truck P = -force acting on truck P by truck Q
∴ Force acting on truck Q by truck P is 719000N backward.
d)
Apply Newton's 2nd law on the motion of truck Q,
F - f = ma = (mv-mu)/0.05
-719000 - 3000 = [(3000)(6.51)-3000u]/0.05
u = 18.5 ms⁻¹
∴ The speed of truck Q before the collision was 18.5 ms⁻¹ to the front.
2015-06-16 21:34:19 補充：
Toi Lam,
If the can travels together with the stone after the collision, asking the velocity of the can after the collision seems stupid...
2015-06-16 21:35:00 補充：
更正：
2. a) ∴ The speed of the trucks after the collision was 6.51 ms⁻¹.
d) The speed of truck Q before the collision was 18.5 ms⁻¹.
（因為 speed 是 scalar，velocity 才是 vector）
2015-06-17 5:25 am
In quwstion 1 , is the can travel together with the stone after collision?
2015-06-16 21:53:26 補充：
polarbearhmh, I haven't look at part (c) yet when I asked , haha
2015-06-16 21:53:26 補充：
polarbearhmh, I haven't look at part (c) yet when I asked , haha
收錄日期: 2021-04-16 16:54:33
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