- #1
Chetlin
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(This isn't homework help, but it's something that's been on my homework (and other places in my mechanics class)...I don't know if it belongs here or in the homework section. If I'm wrong, sorry -_-)
I've had a few questions now in class about a two-dimensional elastic collision between two objects on a frictionless surface, and I noticed something on accident. What I did worked both times but I'm not sure if it will always work.
Let's say you have an object sliding across the surface horizontally at speed v_i. It collides elastically with another object with the same mass. The first object then moves at an angle of 25º (25º above the horizontal) and the other object moves at an angle of -65º (65º below the horizontal). I read that the angles these objects make with each other in these kinds of collisions is always 90º, which may be why this works.
The questions then ask you to find the final speed of each object. If you do the math, break it up into components etc. you get, letting \theta_1 = 25^{\circ} and \theta_2 = -65^{\circ} and v_{1_f} be the final velocity of the first object and v_{2_f} being the final velocity of the second object, v_{2_f} = \frac{v_i}{\cos{\theta_2}-\sin{\theta_2}\cot{\theta_1}} and v_{1_f} = \frac{-v_{2_f} \sin{\theta_2}}{\sin{\theta_1}}. But I noticed just from playing around that v_{1_f} = v_i \cos{\theta_1} and v_{2_f} = v_i \cos{\theta_2} which is much simpler.
Even though I'm a math major it's been a long time since I've done trigonometric identities so I've forgotten many of them, but this looks hard to work with anyway. But does anyone know why this is true (if it even really is)? Does it have something to do with the fact that the angle difference is 90º or something to do with the motion relative to the center of mass?
Thanks a lot :D
I've had a few questions now in class about a two-dimensional elastic collision between two objects on a frictionless surface, and I noticed something on accident. What I did worked both times but I'm not sure if it will always work.
Let's say you have an object sliding across the surface horizontally at speed v_i. It collides elastically with another object with the same mass. The first object then moves at an angle of 25º (25º above the horizontal) and the other object moves at an angle of -65º (65º below the horizontal). I read that the angles these objects make with each other in these kinds of collisions is always 90º, which may be why this works.
The questions then ask you to find the final speed of each object. If you do the math, break it up into components etc. you get, letting \theta_1 = 25^{\circ} and \theta_2 = -65^{\circ} and v_{1_f} be the final velocity of the first object and v_{2_f} being the final velocity of the second object, v_{2_f} = \frac{v_i}{\cos{\theta_2}-\sin{\theta_2}\cot{\theta_1}} and v_{1_f} = \frac{-v_{2_f} \sin{\theta_2}}{\sin{\theta_1}}. But I noticed just from playing around that v_{1_f} = v_i \cos{\theta_1} and v_{2_f} = v_i \cos{\theta_2} which is much simpler.
Even though I'm a math major it's been a long time since I've done trigonometric identities so I've forgotten many of them, but this looks hard to work with anyway. But does anyone know why this is true (if it even really is)? Does it have something to do with the fact that the angle difference is 90º or something to do with the motion relative to the center of mass?
Thanks a lot :D
