phy 07MC No.29

cccc

2007-08-12 8:48 am

tom and john stand on two trolley
both boys are at rest at 'u'
tom=30kg john=27kg
tom hold a ball of 3kg
after tom thrown the ball to john
tom moves backwards with a speed 1ms^-1

What is the speed of john after he has caught the ball

i dont know why the ans is 1.00ms^-1

i calculate 1.11 =.="

回答 (1)

[匿名]

2007-08-12 9:03 am

✔ 最佳答案

Take the forward direction to be positive.

We first find out the velocity of the ball when being thrown out by Tom.

When Tom throws out the ball, by law of conservation of momentum,
30 (0) + 3 (0) = 30 (-1) + 3 (v) (-1 because Tom moves backwards)
v = 10
The velocity of the ball when being thrown out by Tom is 10 ms-1.

When John catches the ball, again, by law of conservation of momentum,
27 (0) + 3 (10) = (27 + 3) (v)
v = 1
Hence, the speed of john after he has caught the ball is 1 ms-1

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