A 1150 kg car traveling initially with a speed of 24.0 m/s in an easterly direct

Question

A 1150 kg car traveling initially with a speed of 24.0 m/s in an easterly direction crashes into the back of a 9 530 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 to the east.

A) What is the velocity of the truck right after the collision? 20.72 M/S

b) What is the change in mechanical energy of the car-truck system in the collision? (Use input values with an adequate number of significant figures to calculate this answer.)

I got part a right but not b.

Explanation / Answer

By conservation of momentum, ( 1150 x 24 ) + (9530 x 20) = ( 1150 x 18 ) + (9530 x v) v= 20.724 towards east change in mech energy = 0.5(1150 (24^2 -18^2) + 9530 (20^2 - 20.724 ^2) =0.5 (289800 - 280984.2) = 4407.7 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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