A 1 300 kg car traveling initially with a speed of 25.0 m/s in an easterly direc

Question

A 1 300 kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the back of a 9 600 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?m/s east
(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.) J
(c) Account for this change in mechanical energy.

Explanation / Answer

a .

m u1 + M u2 = m v1 + M v2 ;

1300 * 25 + 9600 * 20 = 1300 * 18 + 9600 * v2 ;

v2 = 20.947 m /s

b .

initial KE = 0 .5 ( 1300 * 25^2 + 9600 * 20^2 ) = 2.32625 e 6 ;

final KE = 0.5 ( 1300 * 18^2 + 9600* 20.947^2 ) = 2.31672868 e 6

change = - 9.5213168 e 3 J ;

c .

we can account for this energy , the energy before the collison goes into the deforamtion of the bodies of the car and the truck , also some heat losses during the collison

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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