Consider a scenario during which a 2.00kg mass A moving at 2.00m/s strikes anoth
ID: 2091084 • Letter: C
Question
Consider a scenario during which a 2.00kg mass A moving at 2.00m/s strikes another 2.00kg mass B which is at rest. Both are confined to move along the same straight line. Three possible outcomes of the collision that conserve momentum are: a) They stick together and move off at 1.00m/s b) They bounce off one one another and move off in the same direction, one A at 0.50m/s and the other B at 1.50m/s. c) A comes to rest and B moves off at 2.00m/s. In each case the momentum before the collision is: (2.00kg)(2.00m/s)= 4.00 kg*m/s. 1. In each of the three cases above show that momentum is conserved by finding the total momentum after the collision. case a, b, and c) P before= 4.00kg*m/s Case a) P after= ? Case b) P after= ? Case c) P after= ? 2. In each of the three cases, find the kinetic energy lost and characterize the collision as elastic, partially inelastic, or totally inelastic. The kinetic energy before the collision is: KE initial= 1/2(2.00kg)(2.00m/s)^2= 4.00kg*m/s^2= 4.00J Case a, b, and c) KE initial= 4.00J Case a) KE after: ? KE lost? Collision is: ? Case b) KE after: ? KE lost? Collision is: ? Case c) KE after: ? KE lost? Collision is: ? 3. An impossible outcome of such a collision is that A stick to B and they both move off together at 1.414 m/s. First show that this collision would satisfy conservation of kinetic energy and then explain briefly why it is an impossible result.Explanation / Answer
a) intial mometum = m1v1+m2v2 = 2*2 +0 = 4
final momentum = 2*1 + 2*1 = 4
intial mometum = final mometum ,,, hence momentumm conserve
b) final momentum = 2*0.5 + 2*1.5 = 1+3 =4
intial mometum = final mometum ,,, hence momentumm conserve
c) final momentum = 2*0+ 2*2 = 4
intial mometum = final mometum ,,, hence momentumm conserve