p i = m 1 v i = (930 kg)(27.6 m/s) = 2.57 ? 10 4 kg·m/s p f = ( m 1 + m 2 ) v f

Question

pi = m1vi = (930 kg)(27.6 m/s) = 2.57 ? 104 kg·m/s

pf = (m1 + m2)vf = (2410 kg)vf

a) What is the loss of kinetic energy (Ki - Kf) in the situation described in the example. kJ (b) What if the 930 kg car actually moves backwards with a speed of 2.7 m/s right after the collision instead of having a perfectly elastic collision. What is the velocity of the heavier car immediately after the collision? Use the same convention for positive direction as defined in the example. m/s (c) What is the loss of kinetic energy in this case? kJ

Solution The magnitude of the total momentum of the system before the collision is equal to that of only the smaller car because the larger car is initially at rest.

pi = m1vi = (930 kg)(27.6 m/s) = 2.57 ? 104 kg·m/s

After the collision, the mass that moves is the sum of the masses of the cars. The magnitude of the momentum of the combination is the following result.

pf = (m1 + m2)vf = (2410 kg)vf

Equating the initial momentum to the final momentum and solving for vf, the speed of the entangled cars, we have the following.

=  m/s

Explanation / Answer

vf = (2.57*10^4)/2410 = 10.66 m/s

a) loss of KE = ki-kf

ki = 0.5*m1*vi^2 = 0.5*930*27.6^2 = 354.218 kJ

kf = 0.5*(m1+m2)*vf^2 = 0.5*2410*10.66^2 = 136.9 kJ

loss in KE = Ki-Kf = 354.218-136.9 = 217.318 kJ

b)using law of conservation of linear momentum

m1*u1 = -(m1*v1)+(m2*v2)

(930*27.6) = (-930*2.7) + ((2410-930)*v2)

v2 = 19 m/sec

c) loss in KE = ki-kf

ki = 0.5*930*27.6^2 = 354.2 kJ

Kf = (0.5*930*2.7^2) + (0.5*(2410-930)*19^2) = 270.5 kJ

loss in KE = ki-kf = 354.2-270.5 = 83.7 kJ

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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