By accident, a large plate is dropped and breaks into three pieces. The pieces fly apart parallel to the floor, with v1 = 2.75 m/s and v2 = 1.65 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.
m1= kg
m2= kg
To find the masses of pieces 1 and 2, we can use the concept of conservation of momentum. According to the principle of conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.
Let's denote the mass of piece 1 as m1 and the mass of piece 2 as m2.
Before the collision, the plate is in motion parallel to the floor, and its momentum has no component parallel to the floor. This means that the momentum of piece 1 (p1) and piece 2 (p2) must cancel each other out in the horizontal direction.
So, we have:
p1 + p2 = 0
The momentum can be calculated by multiplying the mass (m) of an object by its velocity (v). Therefore, we have:
m1 * v1 + m2 * v2 = 0
Substituting the given values:
m1 * 2.75 m/s + m2 * 1.65 m/s = 0
Now we have one equation with two unknowns. However, we also have another piece of information to utilize. Since the pieces fly apart parallel to the floor after the collision, we can conclude that the magnitude of their velocities remains the same. This implies:
v1 = v2
Substituting this into the equation, we get:
m1 * 2.75 m/s + m2 * 2.75 m/s = 0
Combining like terms:
2.75 m/s (m1 + m2) = 0
From this equation, we can see that the sum of the masses (m1 + m2) must be equal to zero. This means that either m1 or m2 must be negative. However, mass cannot be negative in reality. Therefore, we can conclude that both masses must be zero.
Thus, the masses of pieces 1 and 2 are both zero kg.