soccer, colliding, collision, momentum and so forth

Q: When two people collide in soccer what direction will they move and how fast?

A: This depends on how they collide, their initial velocities, their masses, and the amount of friction. On Tuesday my friends and I played soccer, one of my friends used to referee so he viciously hip-checks people when he steals the ball from them. Said friend weighs around 150 pounds. I weigh 135 pounds. We need to know a few more things before we can find out about the vectors after the collision.

1. The first thing we need to do is convert pounds into kilograms.

160 lb*(4.448 N/lb)*(1/9.80665 kg/N) = (mass of friend)

160 lb*(4.448 N/lb) = 711.68 N

711.68 N*(1/9.80665 kg/N) = 72.571 kg

135 lb*(4.448 N/lb)*(1/9.80665 kg/N) = (mass of me)

135 lb*(4.448 N/lb) = 600.48 N

600.48*(1/9.80665 kg/N) = 61.232 kg.

I remember in a previous post I was 63 kg, that was prior when I started playing sports again.

Anyway.

2. The next thing needed is our instantaneous velocities at the time of collision. I was running about as fast as I do while running 5k, my friend was sprinting pretty fast. I would guess he was running as fast as he would run in a 40-yard dash, it takes him about 5.5 seconds to run 40 yards.

40 yards*(.9144 metres/yards)*(1/5.5 seconds) = (friend’s velocity)

40 yards*(.9144 metres.yards) = 36.58 metres

36.58 metres*(1/5.5 seconds) = 6.65 m/s

5 kilometres*(1000 metres/kilometres)*(1/19 minutes)*(1/60 minutes/seconds) = (my velocity)

5 kilometres*(1000 metres/kilometres) =  5000 metres

5000 metres*(1/19 minutes) = 263.16 meters/minutes

263.16 metres/minutes*(1/60 minutes/seconds) = 4.386 m/s

3. First let us say that we are traveling in the same direction and that we collide elastically. Ignore any momentum that is lost by collisions that we make with the ball and the ground. Momentum is conserved in elastic collisions.

mv = P

P(before collision) = P(after collision)

(friend’s mass)*(friend’s velocity) = P(friend)

72.571 kg*6.65 m/s = 482.6 kg*m/s

(my mass)*(my velocity) = P(me)

61.232 kg*4.386 m/s =  268.56 kg*m/s

P(total) =  751.16 kg*m/s

4. Our kinetic energy is also conserved. Kinetic energy is equal to (p^2)/(2m) or (m/2)(v^2)

(61.232 kg/2)*(4.386 m/s)^2 = KE(me)

30.616 kg*(4.386 m/s)^2 = KE(me)

30.616 kg*19.237 m^2/s^2 = KE(me)

589 J = KE(me)

His kinetic energy is:

(72.571 kg/2)*(6.65 m/s)^2 = KE(friend)

36.286 kg*(6.65 m/s)^2 = KE(friend)

36.286 kg*44.22 m^2/s^2 = KE(friend)

1605 J = KE(me)

KE(total) =  2194 J

5. After the collision the total kinetic energy is the same, and so is the total momentum.

2194 = KE(total)

let ‘(vx)’ be my velocity after the collision

let ‘(vy)’ be my friend’s velocity after the collision

let ‘(mx)’ be my mass

let ‘(my)’ be my friend’s mass

(1/2)(my)(vy)^2 + (1/2)(mx)(vx)^2 = 2194

(my)(vy)^2 + (mx)(vx)^2 = 4388 J

(my)(vy)^2 = 4388 - (mx)(vx)^2

(vy)^2 = (4388 - (mx)(vx)^2)/(my)

(vy) = ((4388 - (mx)(vx)^2)/(my))^(1/2)

(my)(vy) + (mx)(vx) = 751.16 kg*m/s

(my)(vy) = 751.16 kg*m/s - (mx)(vx)

(vy) = (751.16 kg*m/s - (mx)(vx))/(my)

((4388 - (mx)(vx)^2)/(my))^(1/2) = (751.16 kg*m/s - (mx)(vx))/(my)

(vy) = 4.575 m/s

(vx) = 6.845 m/s

6. I move away from the point of collision at 6.845 m/s, my friend moves away from the point of collision at 4.575 m/s in the same direction as me.

Other Assumptions:

There is no friction.

We are being treated as spherical masses as opposed to irregular masses.

The impulse between us is direct.