A stationary curling stone is struck in a glancing collision by a second curling stone of equal mass. If the first stone moves away at a velocity of 0.92 m/s [N71¡ã W] and the second stone moves away at a velocity of 1.25 m/s [N44¡ã E], what was the initial velocity of the second stone?
To find the initial velocity of the second stone, we can use vector addition and subtraction.
First, let's break down the given velocities into their x and y components.
For the first stone:
Velocity = 0.92 m/s [N71° W]
The angle given is measured clockwise from the north, so the x-component will be the southward component and the y-component will be the westward component.
So, the x-component of the first stone's velocity = 0.92 m/s * sin(71°) = -0.848 m/s
The y-component of the first stone's velocity = 0.92 m/s * cos(71°) = -0.456 m/s
For the second stone:
Velocity = 1.25 m/s [N44° E]
The angle given is measured clockwise from the north, so the x-component will be the eastward component and the y-component will be the northward component.
So, the x-component of the second stone's velocity = 1.25 m/s * cos(44°) = 0.898 m/s
The y-component of the second stone's velocity = 1.25 m/s * sin(44°) = 0.970 m/s
Now, let's consider the conservation of linear momentum in the x-direction. Since there are no external horizontal forces acting on the system, the initial x-component momentum is zero. Therefore, the x-component of the initial velocity of the second stone will be equal in magnitude but opposite in direction to the x-component of the first stone's final velocity.
So, the x-component of the initial velocity of the second stone = -(-0.848 m/s) = 0.848 m/s
Next, let's consider the conservation of linear momentum in the y-direction. Again, since there are no external vertical forces acting on the system, the initial y-component momentum is zero. Therefore, the y-component of the initial velocity of the second stone will be equal in magnitude but opposite in direction to the y-component of the first stone's final velocity.
So, the y-component of the initial velocity of the second stone = -(-0.456 m/s) = 0.456 m/s
Finally, we can find the magnitude and direction of the initial velocity of the second stone using the x and y components.
Magnitude of the initial velocity of the second stone = sqrt(0.848^2 + 0.456^2) = 0.970 m/s
To find the direction, we can use trigonometry.
Angle = atan(0.456 m/s / 0.848 m/s) = 28.21°
Since the x-component is positive (eastward) and the y-component is positive (northward), the direction of the initial velocity is [N28° E].
Therefore, the initial velocity of the second stone was 0.970 m/s [N28° E].