To solve this problem, we can use the principle of conservation of momentum.
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Initial Momentum: The initial momentum of the system is only due to the first ball, as it is the only ball moving. The momentum (\(p\)) is given by the formula: \[ p = m \cdot v \] Where \(m\) is mass and \(v\) is velocity.
For the first ball: \[ p_1 = m_1 \cdot v_1 = 0.7 , \text{kg} \cdot 13 , \text{m/s} = 9.1 , \text{kg m/s} \]
The second ball is stationary at this point, thus its initial momentum is 0. Therefore, the total initial momentum of the system is: \[ p_{\text{initial}} = 9.1 , \text{kg m/s} \]
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Collision between the first and second balls: The first ball stops after hitting the second ball, meaning its final momentum is 0. The second ball will then gain all of the momentum that the first ball had (assuming an elastic collision and equal mass for simplicity). So the momentum after the collision (which is now carried by the second ball) becomes: \[ p_{\text{final, second ball}} = p_{\text{initial}} = 9.1 , \text{kg m/s} \] If we denote the velocity of the second ball after the collision as \(v_2\), then: \[ m_2 \cdot v_2 = 9.1 , \text{kg m/s} \] Since \(m_2 = 0.7 , \text{kg}\): \[ 0.7 , \text{kg} \cdot v_2 = 9.1 , \text{kg m/s} \] Solving for \(v_2\): \[ v_2 = \frac{9.1 , \text{kg m/s}}{0.7 , \text{kg}} = 13 , \text{m/s} \]
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Collision between the second ball and the third ball: Now, the second ball (which has a mass of 0.7 kg) collides with the third ball (which has a mass of 0.2 kg). Let’s denote the velocity of the third ball after this collision as \(v_3\).
Before the collision, the momentum of the second ball is given by: \[ p_{\text{initial, second}} = m_2 \cdot v_2 = 0.7 , \text{kg} \cdot 13 , \text{m/s} = 9.1 , \text{kg m/s} \]
The third ball is initially stationary, thus it has: \[ p_{\text{initial, third}} = 0 \]
The total initial momentum before the collision of the second and third ball is: \[ p_{\text{initial, total}} = 9.1 , \text{kg m/s} \]
After the collision, let’s assume that the second ball comes to a stop (which might not be the case if the collision is perfectly elastic). Then the momentum of the system can be described as: \[ p_{\text{final}} = m_2 \cdot 0 + m_3 \cdot v_3 \] So, let’s consider that most momentum transfers: \[ 9.1 , \text{kg m/s} = 0 + 0.2 , \text{kg} \cdot v_3 \] Solving for \(v_3\): \[ v_3 = \frac{9.1 , \text{kg m/s}}{0.2 , \text{kg}} = 45.5 , \text{m/s} \]
Thus, the velocity of the third ball after being hit by the second ball is 45.5 m/s.
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