00:01
If we have a block of mass m on top of a spring, and we drop a block mass of 2m from a height h, a block above block of mass m, we can figure out various parameters.
00:21
First, we're told that this spring is compressed by a distance d from its equilibrium position.
00:26
And we want to know what the spring constant of this spring is.
00:37
We can first to do this for part a using newton's second law.
00:42
We know that the total force is zero, which means that the upward force being the spring force, which should be kx, the downward force, which is the gravity, should be equal.
01:00
So, kd, which is the force of the spring, should be equal to mg.
01:08
Thus, k should be mg over d.
01:12
That's our spring constant.
01:15
For part b, we want an expression for the block b just before it collides with block a.
01:20
We can use conservation of energy.
01:23
That e initial, which is equal to 2mgh, must be equal to e final, which should be equal to 1 .5 .2m .v squared.
01:37
We cancel 2m on both sides.
01:41
We get, and we multiply both sides by 2.
01:45
We get 2gh is v squared.
01:49
So v must be equal to square root 2gh.
01:54
Now, for part c, we want to derive an expression for, or rather for part c, we want to derive an expression for the speed of the blocks immediately after the collision.
02:09
This can be done using conservation of momentum.
02:13
The initial momentum must be equal to the final momentum.
02:17
Initial momentum right before the collision should be 2m times the v that we just got, which is square root 2gh.
02:27
Now, if these two blocks collide inelastically, it means that they move together with the same velocity...