You are on an iceboat on frictionless, flat ice; you and the boat have a combine

Question

You are on an iceboat on frictionless, flat ice; you and the boat have a combined mass M. Along with you are two stones with masses m1 and m2 such that M = 6.0000m1 =12.000m2. To get the boat moving, you throw the stones rearward, either in succession or together, but in each case with a certain speed 20.000 m/s relative to the boat. What is the resulting speed of the boat if you throw the stones simultaneously?

what is the resulting speed of the boat if you throw the stones m1 then m2?

What is the resulting speed of the boat if you throw the stones m2 and then m1?

Explanation / Answer

  In the first case the two stones together have 1/4 the mass of the boat and the rider. So the boat will travel at 1/4 the speed of the rocks. In the other two cases the second stone will travel relative to the earth at a slower speed so the boat will travel at a slower speed.

Boat and person mass = M
stones
m1 = M/6 kg
m2 = M/12 kg
Velocity of stones relative to boat: 20m/s
Velocity of boat: Vb.

MVb = (M/6+M/12)20
Vb = (1/6 + 1/12)20 = 20/4
ANSWER: Vb = 5m/s
momentum of boat: 5M
momentum of stones: (m1 + m2)v = (M/6 M/12)20 = 5M

After throwing first stone, m1. Velocity of boat: Vb1.
(M+m2)Vb1 = m1v = Mv/6
(M+M/12)Vb1 = Mv/6
(13/12)Vb1 = 20/6
Vb1 =20/6(12/13) = 20(1/6)(12/13) = 3.07m/s

After throwing second stone, m2. Velocity of boat: Vb2.
M(Vb2-Vb1) = m2*20
M(Vb2-(2/13)v) = (M/12)20
Vb2-3.07 =20/12
Vb2 =20/12+3.07 =4.73m/s
ANSWER: Vb = 4.73m/s
momentum of boat: 4.73M
momentum of stones: m1v + m2(v-Vb1) = (M/6)v + (M/12)(v-(2/13)v)
= Mv(1/6+(1/12)(11/13)) = Mv(26/156+11/156)=4.73M

After throwing second stone, m1. Velocity of boat: Vb2.
M(Vb2-Vb1) = m1v
M(Vb2-v/14) = (M/6)v
Vb2-20/14 =20/6
Vb2 = 4.76
ANSWER: Vb =4.76m/s
check:
momentum of boat: 4.76M
momentum of stones: m2v + m1(v-Vb1)= (M/12)v + (M/6)(v-V/14)
=4.76M

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