Question

1. A 40.0-kg girl stands on a 15.0-kg wagon holding two 19.5-kg weights. She throws the weights horizontally off the back of the wagon at a speed of 5.0 m/s relative to herself .

A. 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?

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 5.0 m/s relative to herself?

2. A 740.0 N man, carrying a 1.7 kg physics textbook, stands at the center of a frozen pond of radius 5.0 m. He is unable to get to the other side since there is no friction at all between his shoes and the ice. To overcome his difficulty, he throws his textbook horizontally at a speed of 7.0 m/s toward the north shore. How long (in s) does it take him to reach the south shore?

Answer 1

Mass of girl Mg = 40 kg

mass of wagon Mw = 15 kg

mass of first weight m1 = 19.5 kg

mass of second weight m2 = 19.5 kg

A)

initial momentum Pi = 0

after throwing

final momentum Pf = (Mg + Mw)*v1 + (m1+m2)*v

from momentum conservation

Pf = Pi

(40 + 15 )*v1 + (19.5+19.5)*5 = 0

v1 = -3.54 m/s

speed of girl = 3.54 m/s   <<<<--------ANSWER

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B)

initial momentum Pi = 0

after throwing one weight

final momentum Pf = (Mg + Mw + m2)*v1 + (m1)*v

from momentum conservation

Pf = Pi

(40 + 15 + 19.5 )*v1 + (19.5)*5 = 0

v1 = -1.31 m/s

after throwing second weight

momentum before throwing second weight Pi = (Mg + Mw + m1)*v1

momentum after throwing second weight Pf = (Mg + Mw )*v2 + m2*v

Pf = Pi

(40 + 15)*v2 + 19.5*5 = -(40 + 15 + 19.5)*1.31

v2 = -3.55 m/s

speed of girl = 3.55 m/s <<<<--------ANSWER

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2)

weight of man W = 740 N

mass of mas m = W/g = 740/10 = 74 kg

mass of test book mt = 1.7 kg

initial momentum before throwing Pi = 0

final momentum after throwing Pf = m*v + mt*vt

from momentum conservation

Pf = Pi

74*v + 1.7*7 = 0

v = -0.16 m/s

speed of man v= 0.16 towards south shore

time taken to reach south shore t = distance/speed

time t = 5/0.16 = 31.25 s   <<<<--------ANSWER