A Person stands in a stationary canoe and throws a 5kg stone horizontally with a veloctiy of 8m/s. The...?

Question

...Person and canoe have a combined mass of 105kg. (a) Ignoring air resistance and friction, find the velocity of the canoe after the person throws the stone. (b) With what velocity would the person have to throw the stone in order to get the canoe to move at 1m/s?

the answers are (a) 0.38m/s (b)21m/s but i need the work. can someone help?

Answer 1

Okay, the force exerted on the stone will also be exerted on the person on the canoe. But in this case we dont have acceleration, we can still find it by using momentum.

So....

P = mv .... where P is momentum
Pstone = 5kg * 8m/s
Pstone = 40kgm/s

That same momentum will be applied to the canoe and the person that has a combined mass of 105kg.

40kgm/s = mv
40kgm/s = 105
v = .3809 m/s so the answer is (a) 0.38m/s

So if you want the canoe to move 1m/s

Then the momentum applied has to be 105kgm/s

so the velocity the person has to throw the rock is.....

105kgm/s = 5kg(velocity)
velocity = 21 m/s for the canoe and the person to move 1m/s

Answer 2

This problem is an example of coservation of momentum. and also of action and reaction. The approach to the problem is exactly like the "Bullet & recoil " problem.
The momentum of the stone + the momentum of the canoe will be equal to zero. If you consider the velocity of the stone as +ve direction, the backward velocity of the canoe should be taken as -ve.
Mass of the canoe + person = M = 105 kg
Mass of the stone = m = 5 kg
Velocity of the canoe with person = v = ?
Velocity of the stone = V = 8 m/s
According to Newton's third law, when the stone goes forward the canoe will be pushed backwards.
(a) The momntum of the stone = m x V = 40 kg.m/s
The momentum of the canoe = M x -v = - 105 v kg.m/s
By law of conservation of momentum,
mV - Mv = 0
mV = Mv or v = (mV) / M
v = 40 / 105 = 0.381 m/s in the opposite direction.
(b) The rquired velocity of the canoe = v =1 m/s
The mass of canoe and the person = M = 105 kg
The mass of the stone = m = 5 kg
The velocity of the stone = V = ? m/s
Again,
mV - Mv = 0
mV = Mv
V = (Mv) / m
V = ( 105 x 1) / 5 = 21 m/s
The stone should be thrown with a velocity " 21 m/s" so as to push the canoe backwards with a velocity 1 m/s.
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Answer 3

This is a conservation of momentum problem.

Let m = 5 kg stone, M = 105 kg combined mass, and p = mV the momentum of the stone traveling at V = 8 mps. Let P = Mv; the momentum of the combined mass when the stone is tossed.

Then (m + M)v' = p + P = 0; because v' = 0 the initial velocity before the stone was tossed. That is with zero momentum before the toss, the two momenta after the toss have to sum to zero to conserve that zero momentum.

a) From the conservation of momentum, p = mV = -Mv = -P; thus, v = -(m/M)V = -(5/105)8; you can do the math. Note, the minus sign signifies the stone is going one direction and the combo is going in the opposite direction.

b) Use the same p = mV = -Mv = -P; but now V = -(M/m)v; where v = 1 mps. You can do the math.

Answer 4

105 Kg In Stone