Freight car A with a gross weight of 100000 lbs is moving al

Freight car A with a gross weight of 100,000 lbs is moving along the horizontal track in a switching yard at 2 mi/hr. Freight car B with a gross weight of 100:000 lbs and moving at 8 mi/hr overtakes car A and is coupled to it. Determine: the common velocity v of the two cars as they move together after being coupled and the loss of energy |Delta E| due to the impact.

Solution

Momentum before impact is equal to momentum after impact   Total Momentum before imapct = 100000lbs x(8+2) miles /hr Momentum after impact 200000lbs x V= 100000lbs x 10 miles /hr Common velocity after coupling = 100000x10/ 200000= 5 miles / hr                                                                                                                           Kinetic Energy before impact is equal to kinetic energy after impact      1/2 x100000(square of (8x5280/3600)     + square of (2x 5280/ 3600)) =7313500 lb ft square/ sec square                                                                             Kinetic energy after impact = 1/2 x 200000x square of ( 5x5280/3600)

=5377533 lb ft square/sec square  

Hence los of energy after impact 1935967lb ftsq/sec sq =2.624819 e raised to 6 joules