A plate falls vertically to the floor and breaks up into three pieces, which slide along the floor. Immediately after the impact, a 200g piece moves along the x-axis with a speed of 2.00m/s and a 235g piece moves along the the y axis with a speed of 1.50m/s. The third piece has a mass of 100g. In what direction does the third piece move?
I have done this:
tan^-1 = 235(1.50)/200(2) = 41.4
i don't think this is right please help.
Tan^-1(Fy/Fx)
Fx = 200*2
Fy = 235*1.5
I think you are right.
tan^-1 = 235(1.50)/200(2) = 41.4
Angle from x axis= 180 + 41.4
To determine the direction in which the third piece moves, we can use the concept of momentum conservation. According to the law of conservation of momentum, the total momentum before the impact is equal to the total momentum after the impact.
Let's consider the x-axis and the y-axis independently:
On the x-axis:
Mass of the 200g piece = 0.2kg
Velocity of the 200g piece = 2.00m/s
Momentum of the 200g piece = (0.2kg)(2.00m/s) = 0.4 kg·m/s
On the y-axis:
Mass of the 235g piece = 0.235kg
Velocity of the 235g piece = 1.50m/s
Momentum of the 235g piece = (0.235kg)(1.50m/s) = 0.3525 kg·m/s
Now, let's calculate the total momentum before the impact:
Total momentum before = momentum on the x-axis + momentum on the y-axis
Total momentum before = 0.4 kg·m/s + 0.3525 kg·m/s
Total momentum before = 0.7525 kg·m/s
According to the law of conservation of momentum, the total momentum after the impact should be equal to the total momentum before the impact. So, we can write:
Total momentum after = 0.7525 kg·m/s
Now, let's consider the third piece with a mass of 100g (0.1kg). Let's assume its velocity is v_3, and its momentum is p_3.
Total momentum after = momentum of the third piece (p_3)
Total momentum after = p_3
Now, we can equate the total momentum before and the total momentum after:
0.7525 kg·m/s = p_3
Solving for p_3, we find:
p_3 = 0.7525 kg·m/s
Therefore, the third piece moves along the x-axis with a momentum of 0.7525 kg·m/s.
To find the direction in which the third piece moves, we need to analyze the components of the velocities of the other two pieces.
Let's assign positive x to the right and positive y upwards.
Given:
Mass of the first piece (m1) = 200g = 0.2kg
Velocity of the first piece (v1) = 2.00m/s
Mass of the second piece (m2) = 235g = 0.235kg
Velocity of the second piece (v2) = 1.50m/s
Mass of the third piece (m3) = 100g = 0.1kg
To determine the direction of the third piece, we need to calculate the components of velocities of the first two pieces along the x and y axes.
Velocity of the first piece along the x-axis:
v1x = v1 * cos(θ1) (θ1 = angle with the x-axis)
Since the first piece moves along the x-axis, θ1 = 0, and therefore:
v1x = v1 * cos(0) = v1 * 1 = v1
Velocity of the first piece along the y-axis:
v1y = v1 * sin(θ1) (θ1 = angle with the x-axis)
Since the first piece moves vertically downward, θ1 = 180 degrees, and therefore:
v1y = v1 * sin(180) = v1 * 0 = 0
Velocity of the second piece along the x-axis:
v2x = v2 * cos(θ2) (θ2 = angle with the x-axis)
Since the second piece moves along the y-axis, θ2 = 90 degrees, and therefore:
v2x = v2 * cos(90) = v2 * 0 = 0
Velocity of the second piece along the y-axis:
v2y = v2 * sin(θ2) (θ2 = angle with the x-axis)
Since the second piece moves along the y-axis, θ2 = 90 degrees, and therefore:
v2y = v2 * sin(90) = v2 * 1 = v2
Now let's calculate the total momentum along the x-axis and y-axis:
Momentum along the x-axis:
p_x = m1 * v1x + m2 * v2x + m3 * v3x
Since both m1 and v1x are non-zero, the only way to have p_x = 0 is if v3x = 0.
Momentum along the y-axis:
p_y = m1 * v1y + m2 * v2y + m3 * v3y
p_y = m2 * v2 (since v1y = 0 and v3y = 0)
Now we can solve for v3y the following equation:
p_y = m2 * v2 = m3 * v3y
v3y = (m2 * v2) / m3 = (0.235kg * 1.50m/s) / 0.1kg = 3.525m/s
Since v3y is positive and it lies in the positive y-direction, the third piece moves upward along the y-axis.