A 32.0-kg girl stands on a 10.0-kg wagon holding two 12.5-kg weights. She throws

Question

A 32.0-kg girl stands on a 10.0-kg wagon holding two 12.5-kg weights. She throws the weights horizontally off the back of the wagon at a speed of 9.5 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? 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 9.5 m/s relative to herself?

Explanation / Answer

using law of conservation of momentum

momentum before hitting = momentum after hitting

(mb*u1)+(mg*u2) = (mb*v1)+(mg*v2)

(mb*0)+(480*40) = (76*v1)+(mg*v2)

76*v1 = (480*40)-(m*g*v2) = = 0.046*(480*40)

v1 = 0.046*480*40/76 = 11.62 m/sec

using law of conservation of momentum

momentum of the sytem before throwing = momentum of the sytem after throwing

(mg+mw)*v = (2m*v1)

v = (2*m*v1)/(mg+mw) = (2*12.5*9.5)/(32+10) = 5.65 m/sec

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when she throws one at a time

v = (m*v1)/(mg+m+mw) = (12.5*9.5)/(32+10+12.5) = 2.17 m/sec

after throwing second ball also

(mg+mw+m)*v = (mg+mw)*v'-(m*v1)

(32+10+12.5)*2.17 = (32+10)*v'-(12.5*9.5)

v' = 5.64 m/sec

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