Concept explainers
Three identical cars are being unloaded from an automobile carrier. Cars B and C have just been unloaded and are at rest with their brakes off when car A leaves the unloading ramp with a velocity of 5 .76 ft/s and hits car B, which hits car C. Car A then again hits car B. Knowing that the velocity of car B is 5.04 ft/s after the first collision. 0.630 ft/s after the second collision, and 0.709 ft/s after the third collision, determine (a) the final velocities of cars A and C, (b) the coefficient of restitution for each of the collisions.
The final velocities of the car A and C.
Answer to Problem 14.105RP
The value of final velocities of the car A is VA = 0.641ft/s and for car C is Vc=4.41ft/s
Explanation of Solution
Given information:
Based on the principle of impulse momentum system before impact and after impact is same and equal
First consider the momentum of the car A hits car B at first time:
And then consider the momentum when the car B hits car C we get
Formulate the momentum when car A hits car B for second consideration
The coefficient of restitution for each of the collision.
Answer to Problem 14.105RP
The value of coefficient of restitution is e1 = 0.75, e2=0.75, e3=0.75
Explanation of Solution
Here the coefficient of restitution is a difference between relative velocities to the after and before impact.
First, we must formulate the coefficient of restitution when the condition is A hits B for first consideration.
Similarly formulate coefficient of resolution the condition is B hits C, we get:
Formulate the coefficient of resolution then consider A hits B for the second consideration.
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Chapter 14 Solutions
Vector Mechanics For Engineers
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