To solve this problem, we will use principles of conservation of momentum and the concepts related to kinetic energy and impulse.
1. Speed of the tractor immediately after the rope becomes taut
Since the tractor is towing the car, when the rope becomes taut, the system will experience a change in speed due to the interaction between the tractor and the car. Using the conservation of momentum, we can equate the total momentum before and after the rope becomes taut.
Let:
- Mass of the tractor, \( m_t = 5000 , \text{kg} \)
- Mass of the car, \( m_c = 2500 , \text{kg} \)
- Initial speed of the tractor, \( v_t = 3.0 , \text{m/s} \)
- Initial speed of the car, \( v_c = 0 , \text{m/s} \) (since the car is stationary)
Before the rope becomes taut, the total momentum of the system (tractor + car) is: \[ p_{\text{initial}} = m_t \cdot v_t + m_c \cdot v_c = (5000 , \text{kg} \cdot 3.0 , \text{m/s}) + (2500 , \text{kg} \cdot 0) = 15000 , \text{kg m/s} \]
Let \( v_f \) be the final speed of both the tractor and the car after the rope becomes taut. The total mass of the system after the rope becomes taut is: \[ m_t + m_c = 5000 , \text{kg} + 2500 , \text{kg} = 7500 , \text{kg} \]
Using conservation of momentum, we have: \[ p_{\text{initial}} = p_{\text{final}} \] \[ 15000 , \text{kg m/s} = (m_t + m_c) \cdot v_f \] \[ 15000 , \text{kg m/s} = 7500 , \text{kg} \cdot v_f \] \[ v_f = \frac{15000 , \text{kg m/s}}{7500 , \text{kg}} = 2.0 , \text{m/s} \]
2. Loss of kinetic energy of the system just after the car has started moving
The kinetic energy before the rope becomes taut (only the tractor is moving): \[ KE_{\text{initial}} = \frac{1}{2} m_t v_t^2 + \frac{1}{2} m_c v_c^2 = \frac{1}{2} \cdot 5000 , \text{kg} \cdot (3.0 , \text{m/s})^2 + 0 = \frac{1}{2} \cdot 5000 \cdot 9 = 22500 , \text{J} \]
The kinetic energy after the rope becomes taut (both tractor and car move together): \[ KE_{\text{final}} = \frac{1}{2} (m_t + m_c) v_f^2 = \frac{1}{2} \cdot 7500 , \text{kg} \cdot (2.0 , \text{m/s})^2 = \frac{1}{2} \cdot 7500 \cdot 4 = 15000 , \text{J} \]
Now we find the loss in kinetic energy: \[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = 22500 , \text{J} - 15000 , \text{J} = 7500 , \text{J} \]
3. Impulse in the rope when it jerks the car into motion
Impulse can be calculated as the change in momentum experienced by the car. Initially, the car is at rest (momentum = 0) and after the rope becomes taut and pulls the car, it moves with speed \( v_f = 2.0 , \text{m/s} \).
The momentum of the car after being pulled is: \[ p_{\text{car}} = m_c \cdot v_f = 2500 , \text{kg} \cdot 2.0 , \text{m/s} = 5000 , \text{kg m/s} \]
Thus, the change in momentum (impulse) is: \[ \text{Impulse} = \Delta p = p_{\text{car}} - 0 = 5000 , \text{kg m/s} \]
Summary of Results:
- Speed of the tractor immediately after the rope becomes taut: 2.0 m/s
- Loss in kinetic energy of the system just after the car has started moving: 7500 J
- Impulse in the rope when it jerks the car into motion: 5000 kg m/s
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