Physics
posted by Elizabeth .
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 0.094kg 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.

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=p2p1 = 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)