Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Physics 201, Summer 2017 Exam 3, Page 5 of 6 (1500 kg) an accelerate from o to 1

ID: 1637063 • Letter: P

Question

Physics 201, Summer 2017 Exam 3, Page 5 of 6 (1500 kg) an accelerate from o to 110 m/sin only 8.9 seconds. mele 7. A high power r (a) (2 points) Calculate the average power output of the car's engine assuming air resistance is neglibigle. Exp ress your answer in kilowatts 12.25 power wOrle s.a seconds 18539 N (4 points) Once the car reaches its top speed of 110 m/s, the car can go even faster after a small explosion that separates the car into two pieces. After the explosion, the back end of the car (which has a mass of 950 kg) comes to rest and the front end lies forward. What speed does the front end reach after the explosion? (hint: an explosion is a reverse-collision m the perspective of physics. Think about the physics involved during a collision and 300 m pply the same ideas to this problem). mis

Explanation / Answer

a) We know that power is:

P = energy/time

energy = Ke = 1/2 m v^2 = 0.5 x 1500 x 110^2 = 9075000 J

P = 9075000/8.9 = 1019663 W

Hence, P = 1.02 x 10^6 W = 1.02 Mega Watt

b)from the conservation of momentum

m v = (m - m1)V

V = mv/(m - m1) = 1500 x 110/(1500 - 950) = 300 m/s

Hence, V = 300 m/s

c)Let d be the length of the runway.

The amount of work done required to bring the car to rest by the force is:

W = F d

This should be equal to the kinetic energy.

F d = 1/2 (m - m1)V^2

d = 0.5 x (1500- 950) x 300^2/50000 = 495 m

Hence, d = 495 m