Question

Car 1 of mass 3600kg having a speed of 9.5 m/sec collides into car 2 of mass 2000kg traveling at a speed of 3m/sec in the back. a). What is the speed of the center of mass of the system? b). If the two cars stick together what is the speed of the two cars after the combination? c). What is the change in kinetic energy of the system before and after the collision? 4. 5. The "Great American Revolution" roller coaster at Valencia, California includes a loop-the-loop section whose radius is 8 m at the top. What's the minimum speed for a roller-coaster car at the top of the loop if it's to stay on the track?

Answer 1

Solution:

a) velocity of center of mass of the system.

=> Vcm = m1 * v1 + m2 * v2 / m1 + m2

=> Vcm = 3600 * 9.5 + 2000 * 3 / (2000 + 3600)

=> Vcm = 7.17 m/sec

b) Using conservation of momentum.

=> m1 * u1 + m2 * u2 = (m1 + m2) * V

=> 3600 * 9.5 + 2000 * 3 = (3600 + 2000) * V

=> V = 7.17 m/sec

c) Loss in K.E = initial K.E - Final K.E

=> Loss in K.E = 1/2 * m1 * u1^2 + 1/2 * m2 * u2^2 - 1/2 *(m1+m2)V^2

=> Loss in K.E = 1/2 * 3600 * 9.5^2 + 1/2 * 2000 * 3^2 - 1/2 * (2000 + 3600) * 7.178^2

=> Loss in K.E = 27183.68 J

Thanks