If someone could show all the steps on how to answer this that would be awesome,
ID: 1416493 • Letter: I
Question
If someone could show all the steps on how to answer this that would be awesome, I tend to get confused when working with 2-D problems.
Two football players are running along perpendicular paths. The masses and velocities are given in the figure. The small on tackles the larger one at the point represented by the origin of the coordinate system and the two stick together and slide in the mud along the line marked P. The coefficient of friction between the platers and the mud is 0.20. You may assume that the force exerted by the ground on the players' cleats is negligible compared to the huge force they exert on each other. Find the following:
a. The total momentum of the system before the collision (magnitude and direction).
b. The final momentum of the system just after the collision (magnitude and direction). Explain your reasoning.
c. The final velocity of the players (magnitude and direction).
d. The kinetic energy of the system just after the collision.
e. The distance the players slide before coming to rest.
Explanation / Answer
Here,
M = 100 Kg , m = 50 Kg
v1 = 4 m/s
v2 = 6 m/s
a) before the collision
total momentum before the collision = M * v1 * i + m * v2 * j
total momentum before the collision = 100 * 4 i + 50 * 6 j
total momentum before the collision = 400 i + 300 j
total momentum before the collision = sqrt(400^2 + 300^2) at arctan(300/400)
total momentum before the collision = 500 Kg.m/s at 36.9 degree
b)
as the momentum of the system just after the collision is same as the momentum of the system after the collision
total momentum after the collision = 500 Kg.m/s at 36.9 degree
c)
let the final velocity is vf
vf * (50 + 100) = 500 Kg.m/s at 36.9 degree
vf = 3.33 m/s at 36.9 degree with x axis
d)
kinetic energy of the system = 0.5 * 150 * 3.33^2
kinetic energy of the system = 831.7 J
the kinetic energy of the system after the collision is 831.7 J
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