A 48.0- k g girl stands on a 6.0- k g wagon holding two 16.0- k g weights. She t

Question

A 48.0-kg girl stands on a 6.0-kg wagon holding two 16.0-kg weights. She throws the weights horizontally off the back of the wagon at a speed of 9.0 m/s relative to herself . Assuming that the wagon was at rest initially, what is the speed of the girl relative to the ground after she throws both weights at the same time?

I know this is 3.35 m/s

B)Assuming that the wagon was at rest initially, what is the speed relative to the ground with which the girl will move after she throws the weights one at a time, each with a speed of 9.0 m/s relative to herself

Points given for correct answer and detailed explanation.

Explanation / Answer

B. This question can be solved by the law of conservation of momentum.

Before throwing 1st weight:

Momentum, p1=mass of system*0=0

After throwing let the speed of girl be v.

Then, final momentum,p2= (mass of girl+mass of wagon+mass of 1 weight)*v +mass of weight*9

                                                =(48+6+16)v+16*9

Now, p1=p2

0=(48+6+16)v+16*9

v=-2.057 m/s [-ve sign indicates direction is opposite to direction of throw of weight]

Now again for throwing seconde weight.

Momentum, p1=mass of system*(-2.057)=(48+6+16)*(-2.057)=-144

After throwing let the speed of girl be v.

Then, final momentum,p2= (mass of girl+mass of wagon)*v +mass of weight*9

                                                =(48+6)v+16*9

Now, p1=p2

-144=(48+6)v+16*9

v=-5.33 m/s

As speed is asked direction doesn

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