A 31.0- kg girl stands on a 13.0- kg wagon holding two 13.5- kg weights. She thr

Question

A 31.0-kg girl stands on a 13.0-kg wagon holding two 13.5-kg weights. She throws the weights horizontally off the back of the wagon at a speed of 6.0 m/s relative to herself .

1) 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?

2) 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 6.0 m/s relative to herself?

Explanation / Answer

applying the law of conservation of momentum

here mg = 31 kg

mw = 13 kg

mwt = 13.5 kg

vw = 6 m/s

(mg + mw) v = 2 mwt vw

v = 2 * 13.5 *6/(13+ 31)

v = 3.68 m/s

---------------

from the law of conservation of momentum

v1 = mwt V wt /(mg + mw + mt)

v1 = (13.5 *6)/(31 + 13 + 13.5)

V1 = 1.41 m/s

appying the combining effects of first and second ball

final velocity

v2 = mw * vw +(mg +mw +mwt)*v1/(mg+mw)

v2 = ((13 * 6) +(31+ 13+13.5))*1.41/(31+13)

V2 = 4.34 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.