Three identical cars are beingj unloaded from an automobile carrier. Car B and C

Question

Three identical cars are beingj unloaded from an automobile carrier. Car B and C have just been unloaed and are at rest with their brakes off when Car A leaves the unloading ramp with a velocity of 10.5 ft/s and hits Car B, which hits Car C. Car A then again hits Car B knowing that the velocity of Car B is 7.5 ft/s after the first collision, 0.630 ft/s after the second collision and 0.709 ft/s after the third collision, determine:

A) The final velocities of Cars A and C.

B) The coefficient of restitution for each of the collisions.

Image is available via the link below: https://s22.postimg.org/c55b1aybl/145.jpg

Explanation / Answer

after 1st collision

from momentum conservation

momenutm before collision = momentum after collision

mA*vA0 + mB*vB0 = mA*vA1 + m*BVB1

10.5 + 0 = vA1 + 7.5

vA1 = 3 ft/s

after second collison

momentum before collision = momentum after collision

mB*vB1 + mc*vC0 = mB*vB2 + mC*Vc2

7.5 + 0 = 0.63 + vC2

vc2 = 6.87 ft/s

=====================

after third collison

momentum before collision = momentum after collision

mA*vA1 + mB*vB2 = mA*vA3 + mB*vB3

3 + 0.63 = vA3 + 0.709

vA3 = 2.921 ft/s

part(a)

vA3 = 2.921 ft/s    <<<---------answer

part(b)

Vc2 = 6.87 ft/s    <<<---------answer

part(c)

coefficient of restitution for 1st collision = (vB1-Vb0)/(VA0-vA1) = (7.5-0)/(10.5-3) = 1

coefficient of restitution for 2nd collision = (Vc2-vc0)/(vB1-Vb2) = (6.87-0)/(7.5-0.630) = 1

coefficient of restitution for 3rd collision = (VB3-vB2)/(vA1-VA2) = (0.709-0.63)/(3 - 2.921) = 1

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.