Another Conservation of Linear Momentum?

[pinky2468]

So, I am having a hard time getting started on this problem: Please help I have added the picture!

By accident, a large plate is dropped and breaks into three pieces. The pieces fly apart parallel to the floor. As the plate falls, its momentum has only a vertical component and no component parallel to the floor. After the collision, the component of the total momentum parallel to the floor must remain zero, since the net external force acting on the plate has no component parallel to the floor. Using the data shown in the drawing find the masses of pieces 1 and 2.


Any ideas?

 

[pinky2468]

I added the picture if that will help?

 

[wais]

Hi there, it seems like you are working on a conservation of linear momentum problem involving a dropped plate breaking into three pieces. This type of problem can be tricky, but don't worry, I'm here to help!

First, let's review the concept of conservation of linear momentum. This principle states that the total momentum of a system remains constant unless acted upon by an external force. In this case, the external force acting on the plate is only in the vertical direction, so the total momentum in the horizontal direction must remain constant.

To solve this problem, we can use the equation for conservation of linear momentum:

m1v1 + m2v2 + m3v3 = m1v1' + m2v2' + m3v3'

Where m represents mass and v represents velocity. The subscripts 1, 2, and 3 represent the three pieces of the plate before the collision, and the prime symbols (') represent the pieces after the collision.

Since the plate is falling straight down, the initial velocities of all three pieces will be zero in the horizontal direction. This means that the equation becomes:

0 + 0 + 0 = m1v1' + m2v2' + m3v3'

Now, let's look at the given data in the drawing. We know that the total mass of the plate is 1.5kg, and the velocity of piece 3 after the collision is 3m/s. We also know that the pieces fly apart parallel to the floor, meaning that the horizontal velocity of all three pieces will be the same after the collision.

Therefore, we can set up the equation:

0 = m1(3m/s) + m2(3m/s) + m3(3m/s)

Since we are solving for the masses, we can simplify the equation to:

0 = 3(m1 + m2 + m3)

Now, we can use the given masses of pieces 1 and 2 (m1 = 0.5kg and m2 = 0.8kg) to solve for the mass of piece 3:

0 = 3(0.5kg + 0.8kg + m3)

0 = 3(1.3kg + m3)

0 = 3.9kg + 3m3

-3.9kg = 3m3

m3 = -1