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?
2 Answers
See principle below I leave the final math to you.
Explanation:
Work this out as a momentum problem.
Momentum =
Momentum of canoe =
Momentum of rock =
Resolve momentum vector of rock into north and east component:
North component:
East component :
So combined canoe + rock momentum is two vectors:
North:
East :
Resultant vector magnitude is:
Direction:
The new speed is 7.79 m/s and the direction is
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.
where
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
The momentum equation needs to be done separately for the North direction and the East direction.
North:
Entering the data, the momentum equation becomes
Solving for
East:
Entering the data, the momentum equation becomes
Solving for
So this component of the final velocity is actually west.
The new speed is the vector sum of
I hope this helps,
Steve