ssm By accident, a large plate is dropped and breaks into three pieces. The pieces fly apart parallel to the floor, with v1 = 2.95 m/s and v2 = 1.70 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.
What happened to the third piece?
If there were in fact only two pieces and v1 and v2 are opposite directions, as they must be if there are only two pieces, then all you can say is that
M1 v1 = M2 v2
The third piece is said to be 1.30 kg at 3.07 m/s. M1 has an angle of 25 and m2 had an angle of 45 and the question ias aking what are the masses for m1 and m2
To find the masses of pieces 1 and 2, we need to use the principles of conservation of momentum.
First, let's define the directions of motion for the pieces. Since the pieces fly apart parallel to the floor, we can assume that the x-direction is parallel to the floor, and the y-direction is vertical.
The total momentum before the collision is zero in the x-direction, as there is no horizontal component of momentum. In the y-direction, the total momentum before the collision is given by:
P_initial = m1 * v1 + m2 * v2 + m3 * 0
Since the plate falls vertically, its velocity in the y-direction is 0.
We are given v1 = 2.95 m/s and v2 = 1.70 m/s.
Now, let's represent the masses of the three pieces as m1, m2, and m3.
Using the principles of conservation of momentum, we can write:
0 = m1 * 2.95 + m2 * 1.70 + m3 * 0
Since the plate breaks into three pieces, we know that:
m1 + m2 + m3 = total mass of the plate
However, we do not know the total mass of the plate, so we need another equation to solve for the masses of pieces 1 and 2.
We can use the fact that the component of the total momentum parallel to the floor must remain zero after the collision. This means that the momenta in the x-direction of the pieces must cancel each other out. Therefore, we have:
m1 * v1_x + m2 * v2_x + m3 * 0 = 0
Since the pieces fly apart parallel to the floor, their velocities in the x-direction are equal to their initial velocities in the x-direction. Let's assume that the plate was initially at rest, so v1_x and v2_x are both zero.
m1 * 0 + m2 * 0 + m3 * 0 = 0
This equation tells us that the mass of piece 3 (m3) can be any value, as long as pieces 1 and 2 have equal and opposite masses.
To summarize:
m1 + m2 + m3 = total mass of the plate (equation 1)
m1 = m2 (equation 2)
We have two equations and two unknowns (m1 and m2). We can solve this system of equations to find the masses of pieces 1 and 2.
Substituting equation 2 into equation 1:
m2 + m2 + m3 = total mass of the plate
2m2 + m3 = total mass of the plate
Since m3 can be any value, let's assign a value of m3 = M, where M is the total mass of the plate. Then the equation becomes:
2m2 + M = M
Simplifying:
2m2 = 0
Dividing both sides by 2:
m2 = 0
Since m1 = m2, we also have:
m1 = 0
Therefore, the masses of pieces 1 and 2 are both zero.
This implies that the plate did not break into three distinct pieces, which means that there may be some mistake in the given data or assumptions. Please double-check the information or provide additional details if possible.
To find the masses of pieces 1 and 2, we can use the conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
The total momentum before the collision can be calculated using the formula:
p_initial = m1 * v1 + m2 * v2
where m1 and m2 are the masses of pieces 1 and 2, and v1 and v2 are the velocities of the pieces.
We are given the velocities:
v1 = 2.95 m/s
v2 = 1.70 m/s
The total momentum after the collision will be:
p_final = m1 * v1' + m2 * v2'
Since the component of total momentum parallel to the floor remains zero, we know that v1' = v2'.
Using these values, we can rewrite the equation as:
p_final = (m1 + m2) * v1'
Since the total momentum before and after the collision is the same, we can set up the equation:
m1 * v1 + m2 * v2 = (m1 + m2) * v1'
Plugging in the given values:
m1 * 2.95 + m2 * 1.70 = (m1 + m2) * v1'
Now we need to find an additional equation to solve for both masses. We know that the pieces fly apart parallel to the floor, meaning there is no vertical force acting on the plate during the collision. This implies that the vertical component of momentum before and after the collision is zero.
The vertical component of momentum before the collision is m1 * v1 and m2 * v2, and after the collision is m1 * v1' and m2 * v2'.
Setting up the equation:
m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'
Since we know that v1' = v2', we can simplify the equation to:
m1 * v1 + m2 * v2 = (m1 + m2) * v1'
Now we have a system of two equations:
m1 * 2.95 + m2 * 1.70 = (m1 + m2) * v1'
m1 * 2.95 + m2 * 1.70 = m1 * v1 + m2 * v2
We can solve these equations simultaneously to find the masses of pieces 1 and 2.