A large plate falls to the ground and breaks into three pieces. m1 falls with v1=3m/s, 25 degrees to the left from the vertical. m2 falls with v2=1.79m/s, 45 degrees up from the horizon. And m3=1.3kg falls with v3=3.07m/s vertically down.

How would I find m1 and m2?

THank you in advance

4768

To find m1 and m2, we can use the principles of conservation of momentum and conservation of energy.

Step 1: Conservation of Momentum
In this scenario, we can assume that the net external force acting on the plate is negligible (since no external force is mentioned). Therefore, the total momentum before the plate breaks should be equal to the total momentum after the plate breaks.

Let's break down the velocities into their horizontal and vertical components:
v1_horizontal = v1 * cos(25°)
v1_vertical = v1 * sin(25°)

v2_horizontal = v2 * cos(45°)
v2_vertical = v2 * sin(45°)

m1_horizontal = m1 * v1_horizontal
m1_vertical = m1 * v1_vertical

m2_horizontal = m2 * v2_horizontal
m2_vertical = m2 * v2_vertical

To find m1 and m2, we need the equations for conservation of momentum in the horizontal and vertical directions:

Horizontal momentum: m1_horizontal + m2_horizontal = 0
Vertical momentum: m1_vertical + m2_vertical + m3 * v3 = 0

Step 2: Conservation of Energy
The energy conservation principle tells us that the initial energy of the plate (when it's at rest) should be equal to the sum of the energies of the three resulting pieces.

The initial kinetic energy is given by: KE_initial = (1/2) * (m1 + m2 + m3) * (v1^2 + v2^2 + v3^2)

The final kinetic energy is given by: KE_final = (1/2) * (m1 * v1^2 + m2 * v2^2 + m3 * v3^2)

Equating the initial and final kinetic energies, we have KE_initial = KE_final.

Now, we can use the equations from the conservation of momentum and conservation of energy to solve for m1 and m2.

It's important to note that we need additional information, such as the values of v1, v2, v3, and m3, to perform the calculations and find the specific values of m1 and m2.