Please do not cut and paste yahoo answers solutions as your answer. Also if you
ID: 1985056 • Letter: P
Question
Please do not cut and paste yahoo answers solutions as your answer. Also if you could explain in detail it would be appreciated :)A speeding truck, with a mass of 1200 kg, slams into the back of a stationary car,
with a mass of 1000 kg, at a stop light. The two vehicles slide together, eventually coming to rest 40 m from the point of impact. In addition to measuring the length of the skid mark after the collision, the investigating officer measured the static and kinetic coefficients of friction between the wreckage and the road to be 0.45 and 0.35, respectively.
a. What is velocity of the truck and car immediately after the collision?
b. What is the velocity of the truck just before striking the car?
c. How much thermal energy is generated in this collision?
Explanation / Answer
let ' u ' be the velocity of the truck and and car immediately after the collision we have taking g = 9.8 m/s^2 (a) v^2 - u^2 = 2*a*s v = 0 ; u = u ; a = -0.35*9.8 ; s= 40 -u^2 = -2*0.35*9.8*40 u = 16.57 m/s so,the velocity of the truck and and car immediately after the collision is 16.57 m/s (b) applying conservation of momentum let ' v ' be the velocity of the truck before collision 1200*v = (1200+1000)*u 1200*v = 2200*16.57 v = 30.37 m/s so,the velocity of the truck just before striking the car is 30.37 m/s (c) the kinetic energy lost is = 1/2*1200*v^2 - 1/2*(1200+1000)*u^2 = 1/2*1200*30.37^2 - 1/2*2200*16.57^2 = 251350 joules work done against friction = 0.35*9.8*2200*40 = 301840 joules so thermal energy generated = 301840 - 251350 = 50490 joules so,the thermal energy generated = 50.49 kilojoules