Principle of Conservation of Linear Momentum

[helen3743]

I also need help how to start his problem:

"By accident, a large plate is dropped and breaks into three pieces. The pieces fly apart parallel to the floor, with v1 = 3.10 m/s and v2 = 1.60 m/s. 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."

In the drawing, it shows m1 & m2 falling at an angle to the floor but m3 has no angle, and just has a velocity of 3.07 m/s pointing down with a mass of 1.3kg.

How would the formula look like to solve this problem?
I know the normal formula looks like:
x comp & y comp:
m1vf1 + m2vf2 =m1vo1 + m2vo2

I know how to find velocities incorporation the angles, (like doing m1sin25)but I don't know where to fit in the third mass in the formula.

Do you? Thanks!

 

[Chi Meson]

You have two unknowns, m1 and m2.

That means you need two equations. Balance the x-momenta for one equation, and balance the y-momenta for the other.

 

[kyle8921]

The principle of conservation of linear momentum states that in a closed system, the total momentum remains constant before and after a collision. This means that the sum of the momenta of all objects involved in the collision will remain the same. In this case, the plate and its three pieces can be considered a closed system, as there are no external forces acting on it parallel to the floor.

To solve this problem, we can use the formula for conservation of momentum, which is m1v1 + m2v2 = m1v1' + m2v2' + m3v3', where m1 and m2 are the masses of the two pieces before the collision and v1, v2, v1', v2', and v3' are their respective velocities after the collision. We can also assume that the initial velocity of the plate before it breaks is zero, as stated in the problem.

To incorporate the third mass (m3) in the formula, we can rewrite it as m1v1 + m2v2 + m3v3 = m1v1' + m2v2' + m3v3'. Since we are given the velocities of pieces 1 and 2 after the collision, we can substitute those values in the equation. We are also given the velocity of piece 3 (v3) and its mass (m3), so we can solve for the masses of pieces 1 and 2 using algebra.

The formula would look like this:

m1(3.10 m/s) + m2(1.60 m/s) + m3(3.07 m/s) = m1v1' + m2v2' + m3(3.07 m/s)

We can solve for v1' and v2' using the given velocities and the fact that the net external force acting on the plate has no component parallel to the floor. Then, we can plug those values back into the equation and solve for m1 and m2.

I hope this helps you solve the problem. Remember to always consider the principle of conservation of linear momentum when dealing with collisions in a closed system.