# Force of collision

1. Jan 19, 2013

### GeoMeo

I am trying to determine the formula, (which is momentum?) of 2 objects hitting each other head-on. momentum = mass x velocity (2,000 = 200lbs x 10 mph) times 2 for 4,000. If this is correct, then if one of the masses intersected at say a 45 degree angle, what would the formula be for the force (momentum)?
As you can see I am not a physics kind of guy. I am looking for a practical formula for 2 football players colliding, one at an angle, and how it would compare to a straight on collision. My hope is that the one from the angle could achieve equal or more force with less mass but at a higher velocity then a straight on collision.

2. Jan 19, 2013

### mathman

The magnitue stays the same. However, momentum is a vector quantity, so in case of a 45 deg. angle collision, part of the momenta of the cars are in the same direction and part in the opposite direction. Each component has magnitude 1/√2 times the magnitude.

3. Jan 21, 2013

### jbriggs444

The factor of 1/√2 would apply in an ideal situation where, for instance, a race car crumples into a massive teflon-coated concrete barrier at a 45 degree angle. In a collision of equal-sized football players, one would normally model the collision in the center-of-mass frame. That means that there is a half-angle effect at work.

For two players running in the same general direction and converging at a 45 degree angle, each player is running at a 22.5 degree angle relative to the moving center of mass. That means that each player is moving at sin(22.5 degrees) ~= 38% of their running speed relative to the center of mass. Since collision energy scales with the square of relative speed, that's about 15% of the head-on impact energy.

For two players converging at a 135 angle in generally opposite directions, each player is running at a 67.5 degree angle relative to the center of mass. That means that each player is moving at sin(67.5 degrees) ~= 92% of their running speed relative to the center of mass. Square 92% and you get about 85% of the head-on impact energy.

For two players converging at a 90 degree angle they would each be moving at 45 degrees relative to the center of mass. For this situation, the factor of sin(45 degrees) = 1/√2 applies. Square this and you get exactly 50% of the head-on impact energy.

4. Jan 21, 2013

### GeoMeo

jbriggs444, thanks that was very helpful.
So for equal sized football players the impact energy basically is equal for both regardless of angle. That makes sense even I can understand. If player #1 is 200 lbs and running at 10 mphs and the player #2 is 165 lbs and running at 20 mphs, the impact energy at the center-of-mass stays proportionately the same for both at any angle of impact. Correct?

Thus, player #2 doesn't acheive an advantage by colliding with #1 because of angle.
#2 player can only acheive advantage by increasing his speed to the point that his force is greater then the large #1 player. Correct?

5. Jan 21, 2013

### jbriggs444

Yes. For analysis of the collision, individual speeds don't matter, only the relative speed does. But it would be an over-simplification to think that the way the impact energy is dissipated depends only on the player's relative masses.

You are correct that there is no advantage to be had based on angle. But there is also no advantage to be gained by speed. By Newton's third law, the force of the one player on the other is identical to the force of the other player on the one.

A player gains an advantage by arranging matters to that he can tolerate the impact while the opposing player cannot. For instance, by spearing the opponent with helmet on knee.

6. Jan 21, 2013

### roncaassxo_

John exerts enough force on a 30 kg object to cause it to accelerates a 1.75 m/s squared. How much force did he exert? (ignore friction)

7. Jan 22, 2013

### sophiecentaur

Force equals mass times acceleration.
Simples.