Question

A canoe in a still lake is floating North at `8 m/s`. An object with a mass of `20 kg` is thrown North East at `3 m/s`. If the mass of the canoe was `200 kg` before the object was thrown, what is the new speed and direction of the canoe?

Answer 1

See principle below I leave the final math to you.

Explanation:

Work this out as a momentum problem.
Momentum = `m*v`
Momentum of canoe = `200*8 = 1600` kg.m/s directed north
Momentum of rock = `20*83= 60` kg.m/s directed NE ( 45 deg )

Resolve momentum vector of rock into north and east component:

North component: `60 sin45 =60 *0.707 = 42.42`
East component :`60 sin45 =60 *0.707 = 42.42`

So combined canoe + rock momentum is two vectors:

North: `1642.42`
East : `42.42`

Resultant vector magnitude is: `sqrt(1642.42^2 + 42.42^2)`
Direction: `sin^-1(42.42/1642.42)`

Answer 2

The new speed is 7.79 m/s and the direction is `1.56^@` West of North.

Explanation:

Momentum is conserved.
Momentum before = momentum after

The previous equation and the following equation is the statement of conservation of momentum in this case.

`(m_1+m_2)*u = m_1*v_1 + m_2*v_2`
where `m_1 and m_2` are the masses of the canoe and the 20 kg object respectively, u and the 2 v's (`v_1 and v_2`) are the velocities before and after respectively. Since the canoe's course was North and the object was thrown North East, I will break the velocity of the object into North and East components.

`v_"2n" = 3 m/s*sin45 = 2.12 m/s`

The statement of the problem allow 2 different interpretations. When it says that the object was thrown North East at 3 m/s, was the 3 m/s with respect to the canoe or to the lake? If with respect to the lake is what was meant, since the canoe's velocity was 8 m/s, the object was actually thrown South East `-` with respect to the canoe. I will choose to interpret the 3 m/s NE as with respect to the canoe. Therefore I will add the canoe's 8 m/s to the above 2.12 m/s.

`v_"2e" = 3 m/s*cos45 = 2.12 m/s`

The momentum equation needs to be done separately for the North direction and the East direction.

North:
Entering the data, the momentum equation becomes

`(200 kg*+20 kg)*8 m/s = 200 kg*v_"1n" + 20 kg*(8 m/s+2.12 m/s`

Solving for `v_"1n"`

`1760 kg*m/s = 200 kg*v_"1n" + 202.4 kg*m/s`

`200 kg*v_"1n" = 1760 kg*m/s - 202.4 kg*m/s = 1558 kg*m/s`

`v_"1n" = (1558 kg*m/s)/(200 kg) = 7.79 m/s`

East:
Entering the data, the momentum equation becomes

`(200 kg*+20 kg)*0 = 200 kg*v_"1e" + 20 kg*2.12 m/s`

Solving for

`0 = 200 kg*v_"1e" + 20 kg*2.12 m/s`

`200 kg*v_"1e" = - 42.4 kg*m/s`

`v_"1e" = - (42.4 kg*m/s)/(200 kg) = -0.212 m/s`

So this component of the final velocity is actually west.

The new speed is the vector sum of `v_"1n" and v_"1e"`.

`"New speed" = sqrt(7.79^2 + (-0.212)^2) = 7.79 m/s`

`"Direction" = tan^-1(0.212/7.79) = 1.56^@` West of North.

I hope this helps,
Steve