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A woman of mass 45 kg jumps off the bow of a 80 kg canoe that is intially at rest. if her velocity is 3.5 m/s to the right, what is the velocity of the canoe after she jumps?

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Final answer:

To find the velocity of the canoe after the woman jumps, we can use the principle of conservation of momentum. The momentum of the woman must be equal in magnitude and opposite in direction to the momentum of the canoe. By calculating the momentum of the woman and using it to find the velocity of the canoe, we determine that the canoe's velocity after the woman jumps is approximately -1.97 m/s to the left.

Step-by-step explanation:

To solve this problem, we can use the principle of conservation of momentum. The momentum of an object is the product of its mass and velocity. Before the woman jumps off the canoe, the total momentum of the system (woman + canoe) is zero since the canoe is initially at rest. After she jumps, the total momentum of the system is still zero since there is no external force acting on it. Therefore, the momentum of the canoe in the opposite direction must be equal to the momentum of the woman.

Let's calculate the momentum of the woman first. The formula for momentum is:

Momentum = mass * velocity

Momentum = 45 kg * 3.5 m/s = 157.5 kg·m/s

Since the total momentum is zero, the momentum of the canoe in the opposite direction must be -157.5 kg·m/s.

The mass of the canoe is 80 kg, so we can use the formula again to find its velocity:

Mass * Velocity = Momentum

80 kg * Velocity = -157.5 kg·m/s

Velocity = -157.5 kg·m/s / 80 kg = -1.96875 m/s

Therefore, the velocity of the canoe after the woman jumps is approximately -1.97 m/s to the left.

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