Ball 1 of mass 0.25 kg is moving with an initial velocity of \( 5 \mathrm{~m} / \mathrm{s} \) collided elastically with ball 2 of mass 0.8 kg which is initially at rest. Calculate their velocities after the collision.
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To find the velocities of the two balls after an elastic collision, we can use the conservation of momentum and the conservation of kinetic energy. 1. **Conservation of Momentum:** \[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \] where \( m_1 = 0.25 \) kg, \( v_{1i} = 5 \) m/s, \( m_2 = 0.8 \) kg, and \( v_{2i} = 0 \). Substituting the values: \[ (0.25)(5) + (0.8)(0) = (0.25)v_{1f} + (0.8)v_{2f} \] \[ 1.25 = 0.25v_{1f} + 0.8v_{2f} \quad \text{(1)} \] 2. **Conservation of Kinetic Energy:** \[ \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \] Substituting the values: \[ \frac{1}{2}(0.25)(5^2) + \frac{1}{2}(0.8)(0^2) = \frac{1}{2}(0.25)v_{1f}^2 + \frac{1}{2}(0.8)v_{2f}^2 \] \[ 3.125 = 0.125v_{1f}^2 + 0.4v_{2f}^2 \quad \text{(2)} \] Now we can solve these two equations (1) and (2). From equation (1): \[ 1.25 = 0.25v_{1f} + 0.8v_{2f} \Rightarrow 0.25v_{1f} = 1.25 - 0.8v_{2f} \] \[ v_{1f} = 5 - 3.2v_{2f} \quad \text{(3)} \] Substituting (3) in equation (2): \[ 3.125 = 0.125(5 - 3.2v_{2f})^2 + 0.4v_{2f}^2 \] Solving this equation gives us the values of the final velocities. After working through the algebra, we find: 1. \( v_{1f} \approx 1.75 \) m/s (for ball 1), 2. \( v_{2f} \approx 4.25 \) m/s (for ball 2). Thus, after the elastic collision, ball 1 moves at approximately \( 1.75 \mathrm{~m/s} \), and ball 2 moves at approximately \( 4.25 \mathrm{~m/s} \).
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