The white ball in the figure has a speed of 1.67 m/s and the yellow ball is at rest prior to an elastic glancing collision. After the collision the white ball has a speed of 1.43 m/s. To the nearest tenth of a degree, measured counterclockwise from east, what angle does it scatter at if the yellow ball is scattered at 280o?
To find the angle at which the white ball scatters, we need to use the conservation of momentum and the conservation of kinetic energy.
Conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.
In this case, we can write the equation as:
(mass of white ball * initial velocity of white ball) + (mass of yellow ball * initial velocity of yellow ball) = (mass of white ball * final velocity of white ball) + (mass of yellow ball * final velocity of yellow ball)
Let's call the mass of the white ball M1, the initial velocity of the white ball V1i, the final velocity of the white ball V1f, the mass of the yellow ball M2, and the initial velocity of the yellow ball V2i.
We are given:
V1i = 1.67 m/s
V1f = 1.43 m/s
V2i = 0 (since the yellow ball is at rest)
V2f = ? (final velocity of yellow ball)
M2 = M1 (since both balls have the same mass)
Plugging these values into the conservation of momentum equation, we get:
(M1 * 1.67) + (M1 * 0) = (M1 * 1.43) + (M1 * V2f)
Simplifying the equation, we get:
1.67 = 1.43 + V2f
Subtracting 1.43 from both sides, we get:
V2f = 1.67 - 1.43
V2f = 0.24 m/s
Now that we have the final velocity of the yellow ball, we can find the angle at which it scatters.
The angle at which the yellow ball scatters is given as 280 degrees counterclockwise from east. To find the angle counterclockwise from east, we subtract 280 degrees from 360 degrees (a full circle).
360 degrees - 280 degrees = 80 degrees counterclockwise from east.
Therefore, the yellow ball scatters at an angle of 80 degrees counterclockwise from east.
Please note that the scattering angle of the white ball cannot be determined from the given information.