A girl of mass m g is standing on the edge of a motionless boat ( mass m b ) a d

Question

A girl of mass m g is standing on the edge of a motionless boat ( mass m b ) a distance D from shore, as shown in the figure below. She jumps horizontally toward shore with a speed of vj , relative to the boat.

Since she is jumping relative to the boat that is able to move, her speed relative to the shore (it never moves) is:

vg= vj - vb where the vb is the velocity of the boat relative to the shore.

a.What is the initial momentum of the girl:boat system?

b.Assuming there are no horizontal forces external to the girl and the boat, what is the final total momentum of the girl:boat system?

c.Write down the final momentum of the system in terms of the momentum of the boat + the momentum of the girl.

d.Find the horizontal speed of the boat right after the jump .

e. Find the horizontal speed of the girl right after the jump.

Explanation / Answer

mass of boat = mb

mass of girl = mg

so lets fix the frame of reference as the shore ( which is stationary)

we have been told that boat is motionless in the beginning.

so velocity of girl with respect to shore and velocity of boat with respect ot shore ,both are zeroes.

so

a) initial momentum of girl-boat system is zero.

b) given that  there are no horizontal forces external to the girl and the boat . in that condition there should be no chnage in momentum . so finnal momentum should be equal to initial so zero.

c) now inorder to write final momentum equation we need to find the velocity of boat with respect to shore and velocity of boat with respect to shore.

given that initally the velocity of girl is  vj with respect to boat.

at that moment boat also has a velocity with direction opposite to that of girl as vb

therefore velocity of girl with respect to shore =  vg= vj - vb .

momentum = mass * velocity

momentum of girl = mg(vg) = mg( vj - vb).

momentum of boat = mb(- vb) negative because direction is opposite to that of vg.

total momentum =  mg( vj - vb) + mb(- vb) = mg(vj) - 2mb(vb).

d) horizontal speed of boat = -vb .

inorder to find that we use final momentum equation mg(vj) - 2mb(vb).

and we know that final momentum is zero.

so mg(vj) - 2mb(vb) = 0.

therefore vb= mg(vj)/2mb.

horizontal speed of boat = -vb .= - mg(vj)/2mb.

negative symbol implies that the direction is opposite to that of girl. ie., away from the shore.

e) horizontal speed of girl = vj - vb.

again from conservation of momentum we get that mg( vj - vb) + mb(- vb) .

therfore vj - vb = mb(vb) / mg.

horizontal speed of girl = vj - vb. =  mb(vb) / mg.

here it is positive which implies that the direction is towards the shore.

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