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Sports Science Butt Fumble
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<blockquote data-quote="PFanCan" data-source="post: 511444" data-attributes="member: 1989"><p>I am not a physicist, but an engineer. Let me give this a try. It’s been a while, so I am a bit rusty with Engineering 101:</p><p> </p><p>Ok, if there is a mass traveling at a velocity and it crashes head on into an equal mass traveling at the same velocity, the equation is below:</p><p> </p><p>[ATTACH=full]2289[/ATTACH]</p><p> </p><p>Note that this analysis assumes a completely inelastic collision (i.e. no bounce, just stuff getting squished). In this simple case, such as two cars colliding, both cars come to a rest at the point of impact. Essentially, the forces required to stop their momentums are balanced out by each other. Newton’s 3rd law (the one about equal and opposite...) holds, as always.</p><p> </p><p>Now, let’s change the scenario to the same first mass colliding with an immovable wall, as shown below:</p><p> </p><p>[ATTACH=full]2288[/ATTACH]</p><p> </p><p>In this case, the incoming momentum (Momentum = mass * velocity) is the same as with the two masses above. Thus, in order for the wall not to move, it must push back with the same force (force = change in momentum/time elapsed) assuming similar elapsed time of the entire collision. Hence, two cars colliding head on at 50 mph (each) is exactly the same as one car hitting a wall at 50 mph. Jamie was wrong on Mythbusters…</p><p> </p><p>Now, let’s move on to a scenario where the two colliding masses have unequal momentums. In this case, in order to maintain Newton’s third law (forces on both sides of the equation must be equal), any additional force remaining after what is required to stop the momentum of the smaller/slower object is then used to accelerate a mass outward.</p><p> </p><p>[ATTACH=full]2287[/ATTACH]</p><p> </p><p>The force (F2) applied by the larger/faster mass is balanced by the sum of the force (F1) required to stop the incoming momentum and the force used on the outgoing accelerated object (F3).</p><p> </p><p>So, use the case of a bat hitting a ball. The bat’s momentum is much higher than the incoming ball. So, not only does it negate the forward motion of the ball, but also it sends the ball accelerating in the opposite direction.</p><p> </p><p>Note that there is an additional element in the real world with baseballs and bats in that the collision is partially “elastic”. That is, the ball is deformed by the impact elastically. When the ball “bounces” back to its normal spherical shape, this adds even more to the outgoing acceleration. The longer the bat stays in contact with the ball during the swing, the more the elastic “crunch” and the more the elastic “bounce back” to give it more acceleration. This is called "impulse" and is why you follow through on your swing…</p><p> </p><p>In the photo below, look closely and you can see that even the bat is slightly bent backwards!</p><p> </p><p>[ATTACH=full]2286[/ATTACH]</p><p> </p><p>So, this means the following: If a batter just bunts the ball, he is not trying to overcome the momentum of the ball and create any additional acceleration. His goal is to apply just enough force over time to bring the ball's incoming momentum to zero (well, I guess with a little outgoing velocity to send the ball a few feet forward...). In fact, the batter will soften the impact by pulling the bat slightly away—mostly, this will minimize the effect of the elastic bounce-back.</p><p> </p><p>However, if the batter wants to hit a home run, he will swing the bat hard enough to completely stop the ball's incoming momentum and create a huge additional force that accelerates the ball in the other direction.</p><p> </p><p>And that guy’s butt in the OP’s video? After reading the above, I hope you all will agree that a moving butt that causes Sanchez to accelerate away from the impact is actually causing a higher force than a butt that just stops Sanchez in his tracks.</p><p> </p><p>I hope I got everything correct above and we can close the case on Sanchez and his unfortunate collision with a butt.</p><p> </p><p>Anyone still reading this? I think I just put myself to sleep.</p></blockquote><p></p>
[QUOTE="PFanCan, post: 511444, member: 1989"] I am not a physicist, but an engineer. Let me give this a try. It’s been a while, so I am a bit rusty with Engineering 101: Ok, if there is a mass traveling at a velocity and it crashes head on into an equal mass traveling at the same velocity, the equation is below: [ATTACH=full]2289[/ATTACH] Note that this analysis assumes a completely inelastic collision (i.e. no bounce, just stuff getting squished). In this simple case, such as two cars colliding, both cars come to a rest at the point of impact. Essentially, the forces required to stop their momentums are balanced out by each other. Newton’s 3rd law (the one about equal and opposite...) holds, as always. Now, let’s change the scenario to the same first mass colliding with an immovable wall, as shown below: [ATTACH=full]2288[/ATTACH] In this case, the incoming momentum (Momentum = mass * velocity) is the same as with the two masses above. Thus, in order for the wall not to move, it must push back with the same force (force = change in momentum/time elapsed) assuming similar elapsed time of the entire collision. Hence, two cars colliding head on at 50 mph (each) is exactly the same as one car hitting a wall at 50 mph. Jamie was wrong on Mythbusters… Now, let’s move on to a scenario where the two colliding masses have unequal momentums. In this case, in order to maintain Newton’s third law (forces on both sides of the equation must be equal), any additional force remaining after what is required to stop the momentum of the smaller/slower object is then used to accelerate a mass outward. [ATTACH=full]2287[/ATTACH] The force (F2) applied by the larger/faster mass is balanced by the sum of the force (F1) required to stop the incoming momentum and the force used on the outgoing accelerated object (F3). So, use the case of a bat hitting a ball. The bat’s momentum is much higher than the incoming ball. So, not only does it negate the forward motion of the ball, but also it sends the ball accelerating in the opposite direction. Note that there is an additional element in the real world with baseballs and bats in that the collision is partially “elastic”. That is, the ball is deformed by the impact elastically. When the ball “bounces” back to its normal spherical shape, this adds even more to the outgoing acceleration. The longer the bat stays in contact with the ball during the swing, the more the elastic “crunch” and the more the elastic “bounce back” to give it more acceleration. This is called "impulse" and is why you follow through on your swing… In the photo below, look closely and you can see that even the bat is slightly bent backwards! [ATTACH=full]2286[/ATTACH] So, this means the following: If a batter just bunts the ball, he is not trying to overcome the momentum of the ball and create any additional acceleration. His goal is to apply just enough force over time to bring the ball's incoming momentum to zero (well, I guess with a little outgoing velocity to send the ball a few feet forward...). In fact, the batter will soften the impact by pulling the bat slightly away—mostly, this will minimize the effect of the elastic bounce-back. However, if the batter wants to hit a home run, he will swing the bat hard enough to completely stop the ball's incoming momentum and create a huge additional force that accelerates the ball in the other direction. And that guy’s butt in the OP’s video? After reading the above, I hope you all will agree that a moving butt that causes Sanchez to accelerate away from the impact is actually causing a higher force than a butt that just stops Sanchez in his tracks. I hope I got everything correct above and we can close the case on Sanchez and his unfortunate collision with a butt. Anyone still reading this? I think I just put myself to sleep. [/QUOTE]
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