Use the values from PRACTICE IT to help you work this exercise. Suppose the same

Question

Use the values from PRACTICE IT to help you work this exercise. 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 5.05 m/s. If the pickup truck traveling at 18.9 m/s crashes into the compact car, find the following.

(a) the speed of the system right after the collision, assuming the two vehicles become entangled
m/s

(b) the change in velocity for both vehicles

(c) the change in kinetic energy of the system, from the instant before impact (when the compact car is traveling at 5.05 m/s) to the instant right after the collision
KE =  J

Explanation / Answer

Given

mass of car m1 = 8.89*10^2 kg,

mass of truck m2 = 1.76*10^3 kg

velocity of car before collision is u1 = 5.05 m/s, velocity of truck before collision is u2 = 18.9 m/s

a) by conservation of momentum

   m1u1+m2u2 = (m1+m2)V

   V= m1u1+m2u2 / (m1+m2) m/s

   V = (8.89*10^2 *5.05+1.76*10^3*18.9)/(8.89*10^2+1.76*10^3) m/s

   V = 14.252 m/s

b) change in velocity of the car is

   V-u1 = 14.252 - 5.05 m/s = 9.202 m/s

of truck is V - u2 = 14.252-18.9 = -4.648 m/s

c) total k,e of the system before collision is

   0.5*m1u1^2 + 0.5 m2*u2^2 = 0.5*8.89*10^2 *5.05^2 + 0.5*1.76*10^3*18.9^2 J = 325680.66125 J

after collision

   0.5(m1+m1)V^2 = 0.5(8.89*10^2+1.76*10^3)(14.252)^2 J = 269031.7830 J

change in k.e = 325680.66125-269031.7830 =-56648.87825

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