A 1050 kg sports car is moving westbound at 15.0 m/s on a level road when it col
ID: 1447328 • Letter: A
Question
A 1050 kg sports car is moving westbound at 15.0 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.
What is the velocity (magnitude) of the two vehicles just after the collision?
What is the direction of the velocity of the two vehicles just after the collision?
At what speed should the truck have been moving so that it and car are both stopped in the collision?
Find the change in kinetic energy of the system of two vehicles for the situations of part A.
Find the change in kinetic energy of the system of two vehicles for the situations of part C.
Explanation / Answer
A)
Uisng law of conservation of momentum
m1*u1 + m2*u2 = (m1+m2)*V
(6320*14)-(1050*15) =(1050+6320)*V
V = 9.86 m/s
B) direction is along east
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C) M1*u1 = m2*u2
6320*u1 = 1050*15
u1 = (1050*15)/6320 = 2.5 m/s
D) Kf = 0.5*(m1+m2)*V^2 = 0.5*(6320+1050)*9.86^2 = 358254.22 J
Ki = (0.5*m1*u1^2) + (0.5*m2*u2^2) =(0.5*6320*14^2)+(0.5*1050*15^2) = 737485 J
change in KE = Kf-Ki = 358254.22-737485 = -379230.78 J
E) Kf = 0 J
Ki = (0.5*6320*2.5^2)+(0.5*1050*15^2) = 137875 J
Chnage in KE = 137875 J
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