A pickup truck with mass 1.83*10^3 kg and compact car with mass 9.13*10^2 kg are

Question

A pickup truck with mass 1.83*10^3 kg and compact car with mass 9.13*10^2 kg are both traveling eastward, compact car leading the pickup truck. The driver of the compact car slams on the brakes suddenly, slowing the vehicle to 6.20 m/s. If the pickup truck traveling at 18.1 m/s crashes into the compact car, find the following. .(initial speed of car is 14.9 m/s)

a) the speed of the system right after the collision, assuming the two vehicles become entangled

b) the change in velocity for both vehicles

change in vtruck = ? m/s

change in vcar = ? m/s

c) the change in kinetic energy of the system, from the instant before impact (when the compact car is traveling at 6.20 m/s) to the instant right after the collision

change in KE = ? J

Explanation / Answer

mcar = mass of car = 913 kg

mtruck = mass of truck = 183 kg

Vci = velocity of car before collision = 6.20 m/s

Vti = velocity of truck before collision = 18.1 m/s

V = velocity of the combination after collision

a)

Using conservation of momentum ::

mcar Vci + mtruck Vti = (mcar + mtruck) V

913 (6.20) + 183 (18.1) = (913 + 183) V

V = 8.2 m/s

b)

Change in velocity of truck = V - Vti = 8.2 - 18.1 = -9.9 m/s

Change in velocity of car = V - Vci = 8.2 - 6.2 = 2 m/s

c)

initial KE = (0.5) mcar Vci2 + (0.5) mtruck Vti2 = (0.5) (913) (6.20)2 + (0.5) (183) (18.1)2 = 47524.174 J

final KE = (0.5) (mcar + mtruck )V2 = (0.5) (913 + 183) (8.20)2 = 36847.52 J

change in Kinetic energy = 47524.174 - 36847.52 = 10676.654 J

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