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?

Mar 14, 2018

See principle below I leave the final math to you.

Explanation:

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

Resolve momentum vector of rock into north and east component:

North component: $60 \sin 45 = 60 \cdot 0.707 = 42.42$
East component :$60 \sin 45 = 60 \cdot 0.707 = 42.42$

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: ${\sin}^{-} 1 \left(\frac{42.42}{1642.42}\right)$

Mar 16, 2018

The new speed is 7.79 m/s and the direction is ${1.56}^{\circ}$ 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.

$\left({m}_{1} + {m}_{2}\right) \cdot u = {m}_{1} \cdot {v}_{1} + {m}_{2} \cdot {v}_{2}$
where ${m}_{1} \mathmr{and} {m}_{2}$ are the masses of the canoe and the 20 kg object respectively, u and the 2 v's (${v}_{1} \mathmr{and} {v}_{2}$) 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.

${v}_{\text{2n}} = 3 \frac{m}{s} \cdot \sin 45 = 2.12 \frac{m}{s}$

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.

${v}_{\text{2e}} = 3 \frac{m}{s} \cdot \cos 45 = 2.12 \frac{m}{s}$

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 ${v}_{\text{1n}}$

$1760 k g \cdot \frac{m}{s} = 200 k g \cdot {v}_{\text{1n}} + 202.4 k g \cdot \frac{m}{s}$

$200 k g \cdot {v}_{\text{1n}} = 1760 k g \cdot \frac{m}{s} - 202.4 k g \cdot \frac{m}{s} = 1558 k g \cdot \frac{m}{s}$

${v}_{\text{1n}} = \frac{1558 k g \cdot \frac{m}{s}}{200 k g} = 7.79 \frac{m}{s}$

East:
Entering the data, the momentum equation becomes

$\left(200 k g \cdot + 20 k g\right) \cdot 0 = 200 k g \cdot {v}_{\text{1e}} + 20 k g \cdot 2.12 \frac{m}{s}$

Solving for

$0 = 200 k g \cdot {v}_{\text{1e}} + 20 k g \cdot 2.12 \frac{m}{s}$

$200 k g \cdot {v}_{\text{1e}} = - 42.4 k g \cdot \frac{m}{s}$

${v}_{\text{1e}} = - \frac{42.4 k g \cdot \frac{m}{s}}{200 k g} = - 0.212 \frac{m}{s}$

So this component of the final velocity is actually west.

The new speed is the vector sum of ${v}_{\text{1n" and v_"1e}}$.

$\text{New speed} = \sqrt{{7.79}^{2} + {\left(- 0.212\right)}^{2}} = 7.79 \frac{m}{s}$

$\text{Direction} = {\tan}^{-} 1 \left(\frac{0.212}{7.79}\right) = {1.56}^{\circ}$ West of North.

I hope this helps,
Steve