Answer :
Final answer:
The velocity of the canoe after the woman jumps is approximately -0.6866 m/s to the right.
Explanation:
To solve this problem, we can use the principle of conservation of momentum. Since we are dealing with a collision between the woman and the canoe, we can consider the initial momentum of the system to be zero, as the canoe is initially at rest. The final momentum of the system can be calculated by multiplying the mass and velocity of the woman and the canoe respectively.
Using the formula for momentum:
(mass of woman x velocity of woman) + (mass of canoe x velocity of canoe) = 0
Substituting the given values:
(46 kg x 1 m/s) + (67 kg x velocity of canoe) = 0
Simplifying the equation, we find:
67 kg x velocity of canoe = -46 kg x 1 m/s
Solving for the velocity of the canoe:
velocity of canoe = -46 kg x 1 m/s / 67 kg
Therefore, the velocity of the canoe after the woman jumps is approximately -0.6866 m/s to the right.
