A tractor of mass 5.0*10^3kg is used to tow a car of mass 2.5*10^3kg.the tractor moved with a speed of 3.0m/s^1 just before the towing rope became taut.calculate:the speed of the tractor immediately the rope becomes taut.. lost ke of the system just after the car has started moving..impulse on the rope when it jerks the car back into motion
initial momentum = 5*10^3 * 3 = 15 *10^3
final momentum = 7.5*10^3 v
v = 15/7.5 = 2 m/s
initial Ke = .5*5*10^3 * 9 = 22.5 *10^3Joules
final Ke = .5 * 7.5*10^3*4 = 15*10^3 Joules
loss = 7.5*10^3 Joules
impulse = Force*time = change in momentum of car
= 2.5 * 10^3 * 2 = 5*10^3 kg m/s
To answer these questions, we need to use the principles of conservation of momentum and kinetic energy.
1. Speed of the tractor immediately when the rope becomes taut:
Before the rope becomes taut, the tractor is moving with a speed of 3.0 m/s. Since we have no other information about external forces acting on the tractor, we can assume that no external forces, such as friction, are acting on the tractor-car system. According to the principle of conservation of momentum, the total momentum of the system (tractor + car) before and after the rope becomes taut should be the same. Therefore, the speed of the tractor immediately when the rope becomes taut will remain 3.0 m/s.
2. Lost kinetic energy of the system just after the car has started moving:
The kinetic energy of the system before the car starts moving is given by the sum of the kinetic energies of the tractor and the car. The kinetic energy (KE) of an object is given by the formula KE = (1/2)mv^2, where m is the mass of the object and v is its velocity.
KE_1 = (1/2)(mass of tractor)(velocity of tractor)^2 = (1/2)(5.0*10^3 kg)(3.0 m/s)^2 = 22,500 J (joules).
After the car starts moving, the tractor's kinetic energy will decrease, and the car's kinetic energy will increase. The lost kinetic energy (KE_lost) is given by KE_lost = KE_1 - KE_2, where KE_2 is the total kinetic energy of the system after the car starts moving. But since no external forces such as friction are acting on the system, we assume that no energy is lost during the transition, and so KE_lost = 0 J. Therefore, the lost kinetic energy of the system just after the car has started moving is zero.
3. Impulse on the rope when it jerks the car back into motion:
The impulse (J) on an object is equal to the change in momentum (Δp) of that object, which is given by the formula J = Δp = mv2 - mv1, where m is the mass of the object, v2 is its final velocity, and v1 is its initial velocity.
In this case, the rope jerks the car back into motion, so the final velocity (v2) of the car is not given. However, we know that the system's total momentum is conserved. The initial momentum of the system before the rope becomes taut is zero because both the tractor and the car are initially stationary.
Therefore, J = Δp = mv2 - mv1 = (2.5*10^3 kg)(v2) - (2.5*10^3 kg)(0 m/s) = (2.5*10^3 kg)(v2).
So, the impulse on the rope when it jerks the car back into motion is (2.5*10^3 kg)(v2).