1.A 45 kg girl stands on a 7 kg wagon holding two 21 kg weights. She throws the

Question

1.A 45 kg girl stands on a 7 kg wagon holding two 21 kg weights. She throws the weights horizontally off the back of the wagon at a speed of 8 m/s relative to herself.

a) Assuming that the wagon was at rest initially, what's the speed of the girl relative to the ground after she throws both weights at the same time?

b) If the wagon was initially at rest, what's the speed relative to the ground with which the girl will move after she throws the weights one at a time with a speed of 8 m/s each relative to herself?

Explanation / Answer

Perhaps relative to the girl means relative to the girl WHILE she is in motion.

In this case, if the girl's motion relative to the ground is called v, the speed of the weights relative to the ground will be 8 - v. (Because the sum of the velocity of the weights and the girl must equal 8 m/s)

Momentum -->

(8m/s - v)*(21kg + 21kg) = (45 kg + 7 kg)*v

336m/s* kg - 42vkg = 52vkg
336m/s* kg = 94v kg
v = 3.75 m/s

Thus the girl & wagon move with a velocity of 3.75 m/s

B)

Throw off 1st weight
Mass of car + 1weight = 73 kg
Initial momentum = 0
Final momentum = 73 kg * v + -21 Kg * 8 m/s
73 kg * v1 + -168Kg-m/s = 0
73 kg * v1 = 168 Kg-m/s
v1 = 2.30 m/s
After throwing off 1 weight, the velocity of the car and 1 weight = 2.30 m/s forwards.

Now the car and 1 weight have an initial momentum = 73 kg *2.30 m/s = 42 kg-m/s
Throw off 2nd weight

Initial momentum = 42 kg-m/s
Final momentum = 52 kg * vf + -21 kg* 8 m/s
52 kg * vf + -168kg-m/s = 167.9 kg-m/s
52 kg * vf = 335.9 kg-m/s
vf = 6.45 m/s

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