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A girl is skipping stones across a lake. One of the stones accidentally ricochets off a toy boat that is initially at rest in the water (see the drawing below). The 0.094-kg stone strikes the boat at a velocity of 14 m/s, 15° below due east, and ricochets off at a velocity of 10 m/s, 12° above due east. After being struck by the stone, the boat's velocity is 2.2 m/s, due east. What is the mass of the boat? Assume the water offers no resistance to the boat's motion.

Answer will be in kg.

2 answers

Change of the horizontal component of the stone’s linear momentum is
Δp(x) = m•v(0x) – (-mv(1x)} = m{v(ox)+v(1x)} = m{v(0) •cos15 +v1•cos12}
For the boat Δp=p2-p1 = M•u – 0 = M•u.
Thje law of conservation of linear momentum
Δp(x)= Δp
m{v(0) •cos15 +v1•cos12}= M•u.
M = m{v(0) •cos15 +v1•cos12}/u=
=0.094(14•cos15+10•cos12)/2.2 = 1 kg
Ms.Vfx(stone)+Mb.Vf(boat)= Ms.Vox(stone)+Mb.Vo(boat)

Vo(boat)=0 m/s

So, we have:
Ms.Vfx(stone)+Mb.Vf(boat)= Ms.Vox(stone)

Mb.Vf(boat)= Ms.Vox(stone)-Ms.Vfx(stone)

Mb.Vf(boat)=Ms[Vox(stone)-Vfx(stone)]

Mb=Ms[Vox(stone)-Vfx(stone)/Vf(boat)