00:01
Our question says block 1 with asm1 and speed 4 .0 meters per second slides along the x -axis, okay? and the floor is friction in this part, and it undergoes a one -dimensional elastic collision.
00:12
So this tells us what equations we're going to be able to use here with stationary block 2.
00:16
The mass of the two blocks is m2 is equal to 0 .40m1, which are right here.
00:22
And the two blocks slide into a region then after the collision where the coefficient of kinetic friction, mu -sub -k, is equal to 0 .50, where they stop.
00:30
How far into the region will block one and block two slide.
00:33
So it wants us to find this distance.
00:35
So first let's find the final velocity of block one and two, which we can do, considering that they're inelastic collisions.
00:41
This is m1 minus m2 divided by m1 plus m2 times the initial velocity here of block one.
00:57
Well, this is equal to, i'm going to plug in the value for m2.
01:01
So this is 0 .60m1 divided by 1 .4 .4 .2.
01:08
0 .40m1, the m1s cancel, right? and this is still multiplied by the initial velocity of one.
01:17
Okay, plug that value into this expression.
01:20
M1s are gone here.
01:22
This comes out to be 1 .71 meters per second...