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momentum

用戶163188

2011-09-14 12:16 am

4
In a stunt for an action movie, the 100kg actor jumps from a train which is crossing a river bridge. On the river below, the heroine is tied to a raft floating towards a waterfall at 3ms−1. The raft and heroine have a total mass of 200kg.

a
If the hero times his jumps perfectly so as to land on the raft, and his velocity is 12ms−1 at an angle of 80° to the river current, what will be the velocity of the raft immediately after he lands? Draw a vector diagram to show the momentum addition. (Ignore any vertical motion.)

b
If the waterfall is 100m downstream, and the hero landed when the raft was 16m from the bank, would they plummet over the fall? (Assume the velocity remains constant after the hero has landed.)

想問各位(b)點計？？？

回答 (1)

天同

2011-09-14 1:04 am

✔ 最佳答案

(a) Use momentum conservation in direction of river current
200 x 3 + 100 x [12cos(80)] = (100+200)u
where u is the velocity component of the raft in the current directionafter the jump
i.e. u = 2.695 m/s

Use momentum conservation again in direction perpendicular to the river current.
100 x [12sin(80)] = (100+200)v
where v is the velocity component perpendicualr to the river current
i.e. v = 3.939 m/s

Therefore, resultant velocity of the raft after the jump
= square-root[2.695^2 + 3.939^2] m/s = 4.773 m/s
The angle, a say, at which the resultant velociy makes with the current
a = arc-tan[3.939/2.695] degrees = 55.62 degrees

(b) The angle, b say, at which the raft would just not plummet over the fall is
b = arc-tan[16/100] degrees = 9.09 degrees

Clearly, the raft is moving at an angle of 55.62 degres, which is larger than 9.09 degrees. As such, the raft would land on the river bank before reaching the fall.

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