To determine the velocity of the canoe after the woman jumps off, we can use the principle of conservation of momentum.
The total momentum before the jump must equal the total momentum after the jump.
-
Calculate the initial momentum:
- The initial momentum of the system (woman and canoe) is zero because both are at rest. \[ p_{\text{initial}} = (m_{\text{woman}} + m_{\text{canoe}}) \cdot 0 = 0 \]
-
Define the masses:
- Mass of the woman, \( m_{\text{woman}} = 45 \) kg
- Mass of the canoe, \( m_{\text{canoe}} = 80 \) kg
-
Write the momentum after the jump:
- Let \( v_{\text{canoe}} \) be the velocity of the canoe after the woman jumps.
- The momentum after the jump is given by: \[ p_{\text{final}} = m_{\text{woman}} \cdot v_{\text{woman}} + m_{\text{canoe}} \cdot v_{\text{canoe}} \]
where \( v_{\text{woman}} = 3.5 \) m/s to the right (we'll take this direction as positive).
-
Set the initial momentum equal to the final momentum: \[ 0 = (45 , \text{kg} \cdot 3.5 , \text{m/s}) + (80 , \text{kg} \cdot v_{\text{canoe}}) \]
Rearranging the equation to solve for \( v_{\text{canoe}} \): \[ 80 , \text{kg} \cdot v_{\text{canoe}} = - (45 , \text{kg} \cdot 3.5 , \text{m/s}) \]
-
Calculate the right-hand side: \[ 80 , \text{kg} \cdot v_{\text{canoe}} = -157.5 , \text{kg m/s} \]
-
Solving for \( v_{\text{canoe}} \): \[ v_{\text{canoe}} = \frac{-157.5 , \text{kg m/s}}{80 , \text{kg}} = -1.96875 , \text{m/s} \]
This implies that the canoe moves to the left with a velocity of \( 1.97 , \text{m/s} \) (approximately).
Therefore, the final answer for the velocity of the canoe after she jumps, in units of m/s, is: \[ v_{\text{canoe}} \approx -1.97 , \text{m/s} \]
A negative sign indicates it is moving in the opposite direction to the woman.
1 answer