Question

A canoe in a still lake is floating North at `5 m/s`. An object with a mass of `50 kg` is thrown East at `1/4 m/s`. If the mass of the canoe is `300 kg` after the object is thrown, what is the new speed and direction of the canoe?

Answer 1

The canoe's final velocity has magnitude of 5.8335 m/s and direction of `0.41^@` West of North.

Explanation:

The momentum will be conserved, meaning that initial momentum will be equal to the final momentum. Remember that momentum is a vector, so direction of the motion is important.

Let us define the unit vector `hati` to be in the East direction and the unit vector `hatj` to be in the North direction.

The initial momentum is
`p_i = (m_1+m_2)*v_i = 350 kg*5 m/s*hatj = 1750 kg*m/s*hatj`

The final momentum is
`p_f = m_1*v_(f1) + m_2*v_(f2) = 300 kg*v_(f1) + 50 kg*0.25 m/s*hati`

As I said in the first paragraph, initial momentum will be equal to the final momentum. Therefore,
`1750 kg*m/s*hatj = 300 kg*v_(f1) + 50 kg*0.25 m/s*hati`

On the left side of that last equation, the canoe was moving straight North. On the right side of that, the 50 kg object had been thrown East. There was no Eastward motion in the initial condition.

If momentum is to be conserved, the canoe must head on a course slightly to the West of North to cancel out the Eastward momentum of the thrown object. The vector sum of the 2 pieces must be equal to the original momentum.

To analyse this, I suggest that we resolve the velocity `v_(f1)` into its Westward and Northward components `v_(f1N)` and `v_(f1W)`. So that
`v_(f1) = v_(f1N) + v_(f1W)`

The component `v_(f1W)` represents the Westward component of the canoe in the final case. I said above that this is needed to cancel the Eastward momentum of the 50 kg object. Therefore, this must be true
`300 kg*v_(f1W) + 50 kg*0.25 m/s*hati = 0`

Solving that for v_(f1W),
`v_(f1W) = (-50 kg*0.25 m/s*hati)/(300 kg) = -0.0417 m/s`

Now we need `v_(f1N)`. Conservation of momentum says that we still need `1750 kg*m/s*hatj` of momentum heading North. Same value as the initial momentum. That 50 kg mass has no Northward velocity. The canoe's component of momentum heading North, `v_(f1N)`, needs to be `1750 kg*m/s*hatj`.

So we must have
`300 kg*v_(f1N) = 1750 kg*m/s*hatj`
Solving for `v_(f1N)`,
`v_(f1N) = 1750 kg*m/s*hatj/300 kg = 5.8333 m/shatj`

So we have the 2 components of the canoe's velocity being
`v_(f1W) = -0.0417 m/shati` and `v_(f1N) = 5.83333 m/shatj`

The magnitude of the canoe's velocity is
`v_f1 = sqrt(5.8333^2 + 0.0417^2) = 5.8335 m/s`

The direction of the canoe's course is
`arctan(0.0147/5.8333) = 0.41^@` West of North.

Sorry that this got so long. I hope that you can follow the approach. I suggest that you go thru each and every step ... and verify my math!

Steve