00:01
For this problem on the topic of linear momentum, we have in a figure, particle 1 with a mass m1 of 0 .3 kilogram sliding toward the right along the x -axis on a frictionless floor.
00:11
It has a speed of 2 meters per second, and at point x is equal to 0, it undergoes a one -dimensional elastic collision with the stationary particle 2, which has a mass m2 of 0 .4 kilograms.
00:25
Particle 2 reaches a wall at xw, which is at 70 centimeters, bounces from the wall without a losing any speed and we want to know at what position on the x -axis does particle two then collide again with particle one now we'll first find the velocities of the particles after their first collision at x is equal to zero and t is equal to zero so v1 f is equal to m1 minus m2 over m1 plus m2 times the initial speed of one v1 i this is for an elastic collision and this is 0 .3kg minus 0 .4kg divided by 0 .3kg plus 0 .4kg times the initial speed of particle 1, which is 2 meters per second.
01:29
And so the final speed of particle 1 after the first collision is minus 0 .8.
01:36
29 meters per second.
01:40
Similarly, particle 2 has a final speed v2f, that is 2m1 divided by m1 plus m2 times v1 i.
01:53
This is 2 times 0 .3 kg divided by 0 .3 kg plus 0 .4 kg plus 0 .4 kg times 2 meters per second.
02:16
And so the final velocity of 2 is 1 .7 meters per second.
02:23
Now the relative speed of motion, or at the rate of motion of 1 .7 meters per second, we have 2xw to be 140 centimeters...