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 11.000 m/s relative to the boat.

1) What is the resulting speed of the boat if you throw the stones simultaneously?

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

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

Explanation / Answer

Here, M = 6 m1 = 12 m2.
So, Total Mass = M + m1 + m2 = 12m2 + 2m2 + m2 = 15m2.
Velocity of stones relative to boat: v
Velocity of boat: Vb.

a)

15m2Vb = ( 2m2 + m2)v
5Vb = v
Vb = v/5 = 0.20v = 0.2 * 11 = 2.2 m/s

b)

After throwing first stone, m1. Velocity of boat: Vb1.

(M+m2)Vb1 = m1v
(13)Vb1 = 2v

Vb1 = (2/13)v

After throwing second stone, m2. Velocity of boat: Vb2.
M(Vb2-Vb1) = m2v
12m2(Vb2-(2/13)v) = ( m2)v
Vb2-(2/13)v = v/12
Vb2 = v/12+(2/13)v = v(1/12+2/13) = v(13/156+24/156) = 37v/156
Vb = 37v/156 = 0.237v = 0.237*11 =2.607 m/s

c)

After throwing first stone, m2. Velocity of boat: Vb1.
(M+m1)Vb1 = m2v
14m2Vb1 =  m2v
14Vb1 = v
Vb1 = v(1/14)

After throwing second stone, m1. Velocity of boat: Vb2.
M(Vb2-Vb1) = m1v
12m2(Vb2-v/14) = 2m2v
Vb2-v/14 = v/6
Vb2 = v/6+v/14 =v(1/6+1/14) =v(7/42+3/42) = 10v/42
Vb = 5v/21 = 0.238v = 0.238 * 11 = 2.618 m/s

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

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