Al is driving a 1400-kg car at 15 m/s when he hits Bert’s 1700-kg car from behin

Question

Al is driving a 1400-kg car at 15 m/s when he hits Bert’s 1700-kg car from behind. Bert was driving at 8 m/s in the same direction as Al right before the collision. The cars’ bumpers become entangled during the collision so they move as a single object. The coefficient of kinetic friction between the tires and the road is 0.7. What is the total kinetic energy of the cars before the collision? What is the magnitude of the total momentum of the cars before the collision? What is the speed of the wreckage after the collision? How much kinetic energy is lost in the collision? What is the magnitude of the force of kinetic friction on the wreckage?How much work is done by friction on the wreckage? How far does the wreckage slide after the collision?

Explanation / Answer

m1 (Al) = 1400 kg   m2 (Bert) = 1700 kg

speeds before collision

u1 = 15 m/s     u2 = 8 m/s

speeds after collision

v1 (A) = ?        v2 (B) = ?

1)

KEi = 0.5*m1*u1^2 + 0.5*m2*u2^2

KE i = (0.5*1400*15^2)+(0.5*1700*8^2) = 211900 J

2)

initial momentum before collision

Pi = m1*u1 + m2*u2 = (1400*15)+(1700*8) = 34600 kgm/s

3)

after collision final momentum

Pf = (m1 + m2)*v

Pf = Pi

v = Pi/(m1+m2) = 11.16 m/s

(4)

KEf = 0.5*(m1 +m2)*v^2

KEf = 0.5*(1400+1700)*11.16^2 = 193045.68 J

loss = KEi - KE = 18854.32 J

(5)

f = u*(m1+m2)*g = 0.7*(1400+1700)*9.8 = 21266 N

(6)

W = KEf = 193045.68

(7)

W = f*x

x = W/f = 9.1 m

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