A woman of mass 58 kg jumps off the bow of a 88 kg canoe that is intially at rest.
If her velocity is 2 m/s to the right, what is the velocity of the canoe after she jumps?
Answer in units of m/si.
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the momentum before the woman jumps off the canoe is equal to the momentum after she jumps off.
The momentum before the woman jumps off is given by the formula:
momentum = mass * velocity
The momentum of the woman before she jumps off is 58 kg * 2 m/s = 116 kg m/s (to the right).
After the woman jumps off, the momentum of the canoe can be calculated using the formula:
momentum = mass * velocity
The mass of the canoe is 88 kg and the velocity is unknown, so let's call it V. The momentum of the canoe is then 88 kg * V kg m/s.
Since the total momentum before and after the woman jumps off must be the same, we can set up the equation:
116 kg m/s = 88 kg * V kg m/s
To find V, we divide both sides of the equation by 88 kg:
116 kg m/s / 88 kg = V kg m/s
V ≈ 1.32 m/s
Therefore, the velocity of the canoe after the woman jumps is approximately 1.32 m/s to the right.