Dynamics_Part53 - Problem Set: Problem 4. Problem:...

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Unformatted text preview: Problem Set: Problem 4. Problem: Automobiles A and B, which are moving toward each other, suffer a head-on collision. The energy absorbed by each automobile is equal to the vehicle's loss of kinetic energy relative to a moving reference frame whose origin is the center of mass of the two-vehicle system. Mass, velocity, speed and absorbed energy are m, v, v and E, respectively, while subscript A denotes automobile A and subscript B denotes automobile B. (a) Verify that EA /EB = mB /mA . (b) What are EA and EB if mA = 2000 kg, mB = 1000 kg, vA = 126 km/hr and vB = 72 km/hr? Solution: (a) The velocity of the center of mass for this 2-automobile system, v, is given by (mA + mB ) v = mA vA + mB vB Therefore, v is v= mA vA + mB vB mA + mB So, the velocities of the automobiles relative to the center of mass are vA vB mA vA + mB vB mB (vA - vB ) = mA + mB mA + mB mA vA + mB vB mA (vA - vB ) = vB - v = vB - =- mA + mB mA + mB = vA - v = vA - The energies absorbed by the automobiles equal their corresponding kinetic energies prior to the collision, wherefore EA EB = = 1 mA m2 (vA - vB ) (vA - vB ) B mA vA vA = 2 2 (mA + mB )2 1 m2 mB (vA - vB ) (vA - vB ) mB vB vB = A 2 2 2 (mA + mB ) Hence, dividing left-hand and right-hand sides of the equations for EA and EB , we conclude that mB EA = EB mA (b) First, we must convert the automobiles' speeds from km/hr to m/sec. Hence, vA = (126 km/hr)(1000 m/km) = 35 m/sec 3600 sec/hr and vB = (72 km/hr)(1000 m/km) = 20 m/sec 3600 sec/hr ...
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