00:01
For this problem on the topic of center of mass and linear momentum, we are shown in the figure a particle with a given mass sliding rightwards along the x -axis on a frictionless floor with a given speed.
00:14
It reaches the point x is equal to zero at which it undergoes a one -dimensional elastic collision with another stationary particle.
00:22
We are also given the mass of that particle.
00:25
The particle then reaches the wall at the distance xw, it then bounces from the wall with no loss of speed and strikes particle 1.
00:37
We want to know the position on the x -axis that particle 2 will then collide in particle 1.
00:43
Now we know we can find the velocities of the particles after the first collision, which is at x is equal to 0 and t is equal to 0 using the equations we know v1 f.
00:57
The final velocity of particle 1 is m1 minus m2 over m1 plus m2 times the initial velocity of particle 1 and these values are known so this is 0 .3 kg minus 0 .4 kg over 0 .4 kg over 0 .3 k plus 0 .4kg multiplied by the initial velocity of particle 1 which is 2 meters per second.
01:47
So calculating we get the final velocity of particle 1 to be minus 0 .29 meters per second.
01:58
The final velocity of particle 2 b2f is equal to 2 m1 over m1 plus m2 times the initial velocity of particle 1, v1i.
02:20
And this is 2 times 0 .3 kg over 0 .3 kg plus 0 .4kg.
02:38
And this again is multiplied by 2 meters per second.
02:43
And so calculating we get this to be 1 .7 meters per second.
02:48
So now we know the second particle's velocity after the collision as well as the first particle's velocity after the collision.
03:00
And this is after the first collision...