Question

Exercise 8.36 A 1050 kg sports car is moving westbound at 13.0 m/son a level road when it collides with a 6320 kg truck driving east on the same road at 12.0 m/s . The two vehicles remain locked together after the collision.

Part A What is the velocity (magnitude) of the two vehicles just after the collision? v = 8.44   m/s   SubmitMy AnswersGive Up Correct Part B What is the direction of the velocity of the two vehicles just after the collision? to the west to the east SubmitMy AnswersGive Up Correct Part C At what speed should the truck have been moving so that it and car are both stopped in the collision? v = 2.16   m/s   SubmitMy AnswersGive Up Correct Part D Find the change in kinetic energy of the system of two vehicles for the situations of part A. ΔK = −2.81×105   J   SubmitMy AnswersGive Up All attempts used; correct answer displayed Part E Find the change in kinetic energy of the system of two vehicles for the situations of part C. ΔK = 108475   J   SubmitMy AnswersGive Up Incorrect; Try Again; 5 attempts remaining

Answer 1

Take East as the positive direction.
Mass of the truck, M = 6320 kg.
Mass of the car, m = 1050 kg.
Speed of the truck, V = 12 m/s.
Speed of the car, v = 13 m/s.

A)
sing the conservation of momentum,
MV - mv = (M+m)V_final. ...(1)
Substituting, V_final = 8.4 m/s

B)
Since V_final. is positive, it is towards the east.

C)
In equation (1), substituting V_final = 0 (It is at rest after collision)
MV = mv
V = mv/M
   = (1050 x 13) / 6320 = 2.16 m/s.

D)
Kinetic energy after collision = 0.5 x (m+M)V²_final.
= 0.5 x 7370 x 8.4² = 260013.6 J
Kinetic energy before collision = 0.5 x MV² + 0.5 x mv². = 543765 J
Difference in kinetic energy =  Final KE - Initial KE
= 543765 - 260013.6 = -283751.4 J

E)
Here V_final. = 0
So final Kinetic energy = 0 J.
V = 2.16 m/s
Initial Kinetic energy =  0.5 x MV² + 0.5 x mv².
= 0.5 x [6320 x 2.16² + 1050 x 13²] = 103468.296 J
Change in kinetic energy = Final KE - Initial KE
​ = 0 - 103468.296 = -103468.296 J