Car 1 has a mass of m1 = 65 multiplied by 103 kg and moves at a velocity of v01 = +0.95 m/s. Car 2, with a mass of m2 = 92 multiplied by 103 kg and a velocity of v02 = +1.1 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
1 m/s
(b) Determine the velocity of their center of mass after the collision
2 m/s
(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?
To determine the velocity of the center of mass before the collision, we can use the equation:
v_cm = (m1 * v01 + m2 * v02) / (m1 + m2)
Substituting the given values:
v_cm = (65 * 10^3 * 0.95 + 92 * 10^3 * 1.1) / (65 * 10^3 + 92 * 10^3)
= (61.75 * 10^3 + 101.2 * 10^3) / (157 * 10^3)
= 163.95 * 10^3 / (157 * 10^3)
= 1.043 m/s
Therefore, the velocity of the center of mass before the collision is approximately 1.043 m/s (rounded to 3 decimal places).
To determine the velocity of the center of mass after the collision, we can use the conservation of momentum principle:
(m1 * v1_initial + m2 * v2_initial) = (m1 * v1_final + m2 * v2_final)
Since they are coupled after the collision, their final velocity will be the same, denoted as vf.
m1 * v01 + m2 * v02 = (m1 + m2) * vf
Substituting the given values:
65 * 10^3 * 0.95 + 92 * 10^3 * 1.1 = (65 * 10^3 + 92 * 10^3) * vf
61.75 * 10^3 + 101.2 * 10^3 = 157 * 10^3 * vf
163.95 * 10^3 = 157 * 10^3 * vf
Dividing both sides by 157 * 10^3:
vf = 163.95 * 10^3 / (157 * 10^3)
= 1.043 m/s
Therefore, the velocity of the center of mass after the collision is approximately 1.043 m/s (rounded to 3 decimal places).
Comparing the velocities in parts (b) and (c), we can see that they are equal. Therefore, the answer in part (b) is equal to the common velocity (vf) of the two coupled cars after the collision.