To solve this problem, we can use the principle of conservation of momentum. The total momentum before the collision must equal the total momentum after the collision.
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Calculating the initial momentum:
The momentum \( p \) of an object is given by the formula: \[ p = m \times v \] where:
- \( m \) is the mass,
- \( v \) is the velocity.
For the first medicine ball (mass = 11 kg, velocity = 0.8 m/s due west): \[ p_1 = m_1 \times v_1 = 11 , \text{kg} \times 0.8 , \text{m/s} = 8.8 , \text{kg m/s} \text{ (due west)} \]
For the second medicine ball (mass = 11 kg, velocity = 0 m/s since it is at rest): \[ p_2 = m_2 \times v_2 = 11 , \text{kg} \times 0 , \text{m/s} = 0 , \text{kg m/s} \]
Now, we can find the total initial momentum: \[ p_{\text{initial}} = p_1 + p_2 = 8.8 , \text{kg m/s} + 0 , \text{kg m/s} = 8.8 , \text{kg m/s} \]
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Calculating the final momentum:
After the collision, the two medicine balls stick together. Their combined mass \( m_{\text{final}} \) is: \[ m_{\text{final}} = m_1 + m_2 = 11 , \text{kg} + 11 , \text{kg} = 22 , \text{kg} \]
Let \( v_{\text{final}} \) be the final velocity of the two balls together. According to the conservation of momentum: \[ p_{\text{initial}} = p_{\text{final}} \] Therefore, \[ 8.8 , \text{kg m/s} = m_{\text{final}} \times v_{\text{final}} \] Substituting for \( m_{\text{final}} \): \[ 8.8 , \text{kg m/s} = 22 , \text{kg} \times v_{\text{final}} \]
To solve for \( v_{\text{final}} \): \[ v_{\text{final}} = \frac{8.8 , \text{kg m/s}}{22 , \text{kg}} = 0.4 , \text{m/s} \]
Since the first ball was moving due west and the second was at rest, the final velocity of the combined mass will also be due west.
Thus, the final velocity of the two medicine balls after the collision is 0.4 m/s due west.