One billiard ball is shot east at 2.1 . A second, identical billiard ball is shot west at 0.80 . The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90 and sending it north at 1.47 .
To solve this problem, we can apply the principles of conservation of momentum and conservation of kinetic energy.
First, let's analyze the initial motion of the billiard balls:
Ball 1:
- Initial velocity: 2.1 m/s to the east
Ball 2:
- Initial velocity: 0.80 m/s to the west
To find the final velocity of Ball 2 after the collision, we need to consider the conservation of momentum. According to this principle, the total momentum of an isolated system remains constant before and after the collision.
Momentum before the collision:
- For Ball 1: Momentum = mass × velocity = m × 2.1 m/s (since the mass of both balls is the same, we can ignore it for now)
- For Ball 2: Momentum = mass × velocity = m × (-0.80 m/s)
Total momentum before the collision: (m × 2.1 m/s) + (m × (-0.80 m/s))
After the collision, the two balls deflect at right angles, with Ball 2 moving north at 1.47 m/s. To find the final velocity of Ball 2, we can use vector addition. The final velocity of Ball 2 can be viewed as the diagonal of a right-angled triangle, with the horizontal component (-0.80 m/s) and vertical component (1.47 m/s).
Using the Pythagorean theorem, we can find the magnitude of the final velocity of Ball 2:
(final velocity Ball 2)^2 = (-0.80 m/s)^2 + (1.47 m/s)^2
Now, to find the direction of the final velocity of Ball 2, we can use trigonometry. The angle between the initial velocity of Ball 2 and the final velocity of Ball 2 is given as 90°. Therefore, the angle between the final velocity of Ball 2 and the north direction is:
angle = atan(1.47 m/s / 0.80 m/s)
Finally, with the magnitude and angle of the final velocity of Ball 2 determined, we have solved the problem.