3. (24 points) A speeding truck, with a mass of 1200 kg, slams into the back of
ID: 1983152 • Letter: 3
Question
3. (24 points) 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.Please show all work and explain answers, I have a final coming up! Thanks :]
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
a: The first part is work & energy. Since 'the two vehicles slide together', we use a single common velocity, vtc. Kinetic energy of truck and car - work done by friction = 0 (everything at rest) 1/2(mt+mc)*vtc^2 - mu.kinetic * normal force * slide distance = 0 The normal force is the mass of the car and truck multiplied by gravity (i.e. (mt+mc)*g.) We choose the coefficient of kinetic friction because the wreck is moving relative to the road. Rearranging and solving for vt: vt = sqrt(2 * mu.kinetic * g * slide distance) = sqrt(2 * 0.35 * 9.81 * 40) = 16.57m/s b: From the point immediately before collision to immediately after, linear momentum is conserved: mt*vt1 + mc*vc1 = (mt+mc)*vtc Since the car is initially stationary, vc1 = 0 Rearranging and solving for vt1: vt1 = (mt+mc)*vtc/mt = (1200kg + 1000kg) * 16.57m/s / 1200 kg = 30.38 m/s c: The thermal energy generated in the collision is a result of the work done by friction. W = mu.kinetic*mtc*g*slidedistance = 0.35 * (1000 +1200)kg * 9.81m/s^2 * 40m = 302148J
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