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 13.000 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 and then m2?

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

Explanation / Answer

M = 6.0000m1 =12.000m2.
then
m1= 2m2

We have to consider conservation of momentum.
m1u1 +m2u2=m1v1 +m2v2

a) M * Vb = (M / 6 + M / 12) *v
Vb = (1 / 6 + 1 / 12) v = v / 4
Vb = v / 4

Vb = 13 / 4 = 3.25 m/s
the speed of the boat when the stone thrown simultaniously is 3.25 m/s

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

Conservation of momentum (Keep in mind that the m2 remains on board the boat)
(M + m2) vM -m1v1=0

then   vM = m1v1 /(M + m2 )
vM = m1 12 /(6.00m1 + 0.5m1)
vM = 13/ 6.5= 2 m/s
then we repeat for m2
(M+   m2 ) vM - M v'M    - m2v2 =0
vM' =[(m2v2 + (M+   m2 ) vM] / M
vM' = [(0.5m1 13+ (6m1+   0.5m1) 2] / 6m1

vM' = [(6.5 + (6.5 )2] / 6

      = 3.25 m/s

vM'=3.25 m/s

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

This is a bit simpler.
M vM - (m1+ m2 ) v1=0
vM = [(m1 + m2 ) / M ] v1
vM= [(2m2 + m2 ) / 12m2] 13
vM = [(3 ) / 12 ] 13
vM = 3.25 m/s

Law of conservation of momentum is confirmed since we have the same speed of the boat as in above part.

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

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