Car 1 has a mass of m1 = 65 103 kg and moves at a velocity of v01 = +0.76 m/s. Car 2, with a mass of m2 = 92 103 kg and a velocity of v02 = +1.5 m/s, overtakes car 1 and couples to it. Neglect the effects of friction in your answer.
(a) Determine the velocity of their center of mass before the collision.
(b) Determine the velocity of their center of mass after the collision
(c) Should your answer in part (b) be less than, greater than, or equal to the common velocity vf of the two coupled cars after the collision? Justify
I figured out a and b, but I can't seem to figure out C.
The velocity of the center of mass should not change, because no outside forces act (other than weight, which remains balanced by the upward force of the road.) They tell you to neglect friction. Friction (such as braking during impact) could alter the CM velocity.
The numbers used for the masses of the cars and their velocity are preposterous. They are as massive as houses and not much faster than turtles.
For part c, it says regarding the common velocity Vf isn't that different than Vcm?
if the cars are hooked, the common velocity has to be the velocity of the center of mass.
If they have a common velocity, it must be the CM velocity.
The formula for the final velocity, assuming conservation of momentum, is the same as the formula for the initial CM velocity. If V1 and V2 are initial velocities,
Vcm = [V1*M1 + V2*M2]/(M1 + M2)
(M1+M2)Vf = M1*V1 + M2*V2
Vf = [V1*M1 + V2*M2]/(M1 + M2)
To solve this problem, we need to use the principle of conservation of momentum. According to this principle, the total momentum of a system remains constant unless external forces act on it.
(a) To determine the velocity of the center of mass before the collision, we need to find the total momentum of the two cars individually.
The momentum, p1, of Car 1 is given by:
p1 = m1 * v01
Substituting the given values:
p1 = (65 * 10^3 kg) * (+0.76 m/s)
p1 = 49,400 kg*m/s
Similarly, the momentum, p2, of Car 2 is given by:
p2 = m2 * v02
Substituting the given values:
p2 = (92 * 10^3 kg) * (+1.5 m/s)
p2 = 138,000 kg*m/s
The total momentum before the collision is the sum of the individual momenta:
p_total = p1 + p2
p_total = 49,400 kg*m/s + 138,000 kg*m/s
p_total = 187,400 kg*m/s
The velocity of the center of mass before the collision is given by dividing the total momentum by the total mass of the system:
v_cm_before = p_total / (m1 + m2)
Substituting the given values:
v_cm_before = 187,400 kg*m/s / ((65 * 10^3 kg) + (92 * 10^3 kg))
v_cm_before ≈ 0.948 m/s
Therefore, the velocity of their center of mass before the collision is approximately 0.948 m/s.
(b) After the collision, Car 1 and Car 2 form a system with a common velocity, v_f. Since we are neglecting the effects of friction, we can assume that the total momentum of this system remains constant.
Using the principle of conservation of momentum again, the total momentum after the collision is given by:
p_total = (m1 + m2) * v_f
We already know the total momentum before the collision is 187,400 kg*m/s. So, we can equate these two values:
187,400 kg*m/s = (m1 + m2) * v_f
Solving for v_f:
v_f = 187,400 kg*m/s / (m1 + m2)
Substituting the given values:
v_f = 187,400 kg*m/s / ((65 * 10^3 kg) + (92 * 10^3 kg))
v_f ≈ 1.07 m/s
Therefore, the velocity of their center of mass after the collision is approximately 1.07 m/s.
(c) The answer in part (b) should be equal to the common velocity, v_f, of the two coupled cars after the collision. This is because the principle of conservation of momentum states that the total momentum of a system remains constant unless external forces act on it. In this case, neglecting frictional forces, there are no external forces acting on the system after the collision. Thus, the two cars, when coupled, will move with a common velocity that is equal to the velocity of the center of mass after the collision.
Hence, the answer in part (b) is equal to the common velocity, v_f, of the two coupled cars after the collision.