A 1050 kg sports car is moving westbound at 14.0 m/s on a level road when it col
ID: 1500607 • Letter: A
Question
A 1050 kg sports car is moving westbound at 14.0 m/s on a level road when it collides with a 6320 kg truck driving east on the same road at 14.0 m/s . The two vehicles remain locked together after the collision.
(A) What is the velocity (magnitude) of the two vehicles just after the collision?
(B) What is the direction of the velocity of the two vehicles just after the collision?
(C) At what speed should the truck have been moving so that it and car are both stopped in the collision? (ANS in m/s please)
(D) Find the change in kinetic energy of the system of two vehicles for the situations of part A.
(E) Find the change in kinetic energy of the system of two vehicles for the situations of part C.
Explanation / Answer
m1 = 1050 m2 = 6320 kg
u1 = -14 m/s u2 = +14 m/s
from momentum before conservation
momentum before collision = momentum after collision
m1*u1 + m2*u2 = (m1+m2)*V
-(1050*14)+(6320*14) = (1050+6320)*V
v = 10.01 m/s
(B)
both the vehicles are moving towards east
(c)
final speed of both the cars = 0
from momentum before conservation
momentum before collision = momentum after collision
m1*u1 + m2*u2 = (m1+m2)*V
-(1050*14)+(6320*u2) = (1050+6320)*0
u2 = 2.35 m/s <<<-answer
(D)
dK = Kf - Ki
dK = 0.5*(m1+m2)*v^2 - (0.5*m1*u1^1 + 0.5*m2*u2^2)
dK = (0.5*(1050+6320)*10.01^2)-(0.5*1050*14^2)-(0.5*6320*14^2)
dK = -353022.6315 J
E)
dK = Kf - Ki
dK = 0 - (0.5*1050*14^2)-(0.5*6320*2.35^2)
dK = -120351.1
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