? v truck = m/s ? v car = m/s An pickup truck with mass1.83 103 kg is traveling

Question

?vtruck = m/s ?vcar = m/s An pickup truck with mass1.83  103 kg is traveling eastbound at +15.4 m/s, while a compact car with mass 9.29  102 kg is traveling westbound at -15.4 m/s. (See figure.) The vehicles collide head-on, becoming entangled. Find the speed of the entangled vehicles after the collision m/s Find the change in the velocity of each vehicle. Find the change in the kinetic energy of the system consisting of both vehicles. Suppose the same two vehicles are both traveling eastward, the compact car leading the pickup truck. The driver of the compact car slams on the brakes suddenly, slowing the vehicle to 6.13 m/s. If the pickup truck traveling at 17.0m/s crashes into the compact car, find the following. the speed of the system right after the collision, assuming the two vehicles become entangled m/s the change in velocity for both vehicles the change in kinetic energy of the system, from the instant before impact (when the compact car is traveling at 6.13 m/s) to the instant right after the collision?KE = J

Explanation / Answer

a) conservation of momentum

1.83E3*15.4 - 9.29E2*15.4 = (1.83E3 + 9.29E2)*v

v=5.03 m/s

b) dvtruck = 5.03-15.4=-10.37

dvcar = 5.03 - (-15.4)= 20.43

c)

dKE = KEfinal - KE initial = 1/2*(1.83E3+9.29E2)*5.03^2 - 0.5*1.83E3*15.4^2 - 0.5*9.29E2*15.4^2
=--292260 J


2) a) conservation of momentum

1.83E3*17 + 9.29E2*6.13 = (1.83E3 + 9.29E2)*v
v= 13.34

b) dvtruck = 13.34 - 17=-3.66

dvcar = 13.34 - 6.13=7.21

c) dKE = Ke final - Ke intial

=1/2*(1.83E3+9.29E2)*13.34^2 - 0.5*1.83E3*17^2 - 0.5*9.29E2*6.13^2
=-36400 J

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