A 1440 kg car traveling initially with a speed of 25.0 m/s in an easterly direct

Question

A  1440 kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a 9000-kg truck moving in the same direction at 20.0 m/s. The velocity of the car right after the collision is 18.0 m/s  in its original direction . a) What is the speed (m/s)  of the truck right after the collision? b) How much mechanical energy (J) is lost  in the collision Hint use all the digits for the answer to part a in calculating the answer in part b dont round it off to three figures before using it?

Explanation / Answer

using conservation of linear momentum,

m1u1 + m2u2 = m1v1 + m2v2

=> 1440*25 + 9000*20 = 1440*18 + 9000*v2

so v2 = 21.12 m/s in easterly dirn.

energy before collision = 1/2*m1*u1^2 + 1/2*m2*u2^2

=> E1 = 0.5*1440*25^2 + 0.5*9000*20^2 = 2250000 J

energy after collision = 1/2*m1*v1^2 + 1/2*m2*v2^2

=> E2 = 0.5*1440*18^2 + 0.5*9000*21.12^2 = 2240524.8 J

so energy lost = E1 - E2 = 9475.2 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.