To solve the problem of the elastic collision between the two carts, we can use the conservation of momentum and the conservation of kinetic energy.
Let:
- \( m_1 = 5 , \text{kg} \) (mass of the first cart)
- \( m_2 = 1 , \text{kg} \) (mass of the second cart)
- \( u_1 = 0 , \text{m/s} \) (initial velocity of the first cart)
- \( u_2 = 3 , \text{m/s} \) (initial velocity of the second cart)
Let:
- \( v_1 \) be the final velocity of the first cart
- \( v_2 \) be the final velocity of the second cart
Step 1: Conservation of Momentum
The total momentum before the collision must equal the total momentum after the collision. \[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \] Substituting the known values: \[ 5 \cdot 0 + 1 \cdot 3 = 5 v_1 + 1 v_2 \] This simplifies to: \[ 3 = 5 v_1 + v_2 \quad \text{(1)} \]
Step 2: Conservation of Kinetic Energy
The total kinetic energy before the collision must equal the total kinetic energy after the collision. \[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \] Substituting the known values: \[ \frac{1}{2} \cdot 5 \cdot 0^2 + \frac{1}{2} \cdot 1 \cdot 3^2 = \frac{1}{2} \cdot 5 v_1^2 + \frac{1}{2} \cdot 1 v_2^2 \] This simplifies to: \[ \frac{1}{2} \cdot 1 \cdot 9 = \frac{1}{2} \cdot 5 v_1^2 + \frac{1}{2} \cdot v_2^2 \] \[ 4.5 = \frac{5}{2} v_1^2 + \frac{1}{2} v_2^2 \] Multiply everything by 2 to eliminate the fraction: \[ 9 = 5 v_1^2 + v_2^2 \quad \text{(2)} \]
Step 3: Solving the System of Equations
Now we have a system of two equations:
- \( 3 = 5 v_1 + v_2 \)
- \( 9 = 5 v_1^2 + v_2^2 \)
From equation (1), we can express \( v_2 \) in terms of \( v_1 \): \[ v_2 = 3 - 5 v_1 \]
Substituting \( v_2 \) into equation (2): \[ 9 = 5 v_1^2 + (3 - 5 v_1)^2 \] Expanding the square: \[ 9 = 5 v_1^2 + (9 - 30 v_1 + 25 v_1^2) \] Combining the terms: \[ 9 = 30 v_1^2 - 30 v_1 + 9 \] Subtracting 9 from both sides: \[ 0 = 30 v_1^2 - 30 v_1 \] Factoring out \( 30 v_1 \): \[ 0 = 30 v_1(v_1 - 1) \] This gives two solutions: \[ v_1 = 0 \quad \text{or} \quad v_1 = 1 \]
Step 4: Finding \( v_2 \)
If \( v_1 = 0 \), then substituting into (1): \[ v_2 = 3 - 5(0) = 3 \] This is the initial condition, so we discard this solution.
If \( v_1 = 1 \): \[ v_2 = 3 - 5(1) = 3 - 5 = -2 \]
Final Answer
Thus, the velocity of the second cart after the collision is: \[ \boxed{-2 , \text{m/s}} \]