About momentum

2009-11-01 7:57 am

A box truck P of mass 5500kg is at rest. Another box truck Q of mass 3000kg collides with truck P and both trucks move forwards for 30m after collision.
Assume the friction acting on each truck = 3000N during collision
and the friction acting on the trucks = 6000N when they moved together.
(a) What was the speed of the trucks after the collision ?
(b) If the time of the collision was 0.05s, 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,

回答 (1)

Prof. Physics

2009-11-01 8:04 am

✔ 最佳答案

a. After collision:

By Newton's law of motion,

F = (M + m)a

-6000 = (5500 + 3000)a

Acceleration, a = -0.706 ms^-2

By equation of motion, v^2 = u^2 + 2as

0 = u^2 + 2(-0.706)(30)

Speed after collision, u = 6.51 ms^-1

b. By impulse = Change of momentum

Ft = Mv - Mu

F(0.05) = (5500)(6.51 - 0)

Force, F = 7.16 X 10^5 N

c. By Newton's 3rd law of motion, the force acting on P by Q and that on Q by P are action-reaction pair, so they have the same magnitude but they are opposite in direction.

So, the required force = -7.16 X 10^5 N

d. By impulse = change of momentum

Ft = mv - mu

(-7.16 X 10^5)(0.05) = (3000)(6.51 - u)

Initial speed, u = 18.4 ms^-1

參考: Physics king

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