Final answer:
The final velocity of the second billiard ball after the collision is 2.679 m/s to the right, calculated using conservation of momentum.
Step-by-step explanation:
The question involves finding the final velocity of a billiard ball after a collision, which is a common problem in physics, specifically in the topic of momentum and collisions. We can use the principles of conservation of linear momentum because the masses of the balls are equal, and the collision can be assumed to be perfectly elastic.
To solve this, we first need to use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision. Since the balls have equal masses and the first ball hits the second ball which is initially at rest, we can set up the equation:
m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final
where m1 and m2 are the masses of ball 1 and ball 2 respectively, and v represents their velocities. Given that ball 1's final velocity is at an angle, we must resolve its velocity into horizontal and vertical components.
To find the horizontal velocity of ball 1 after the collision, we use the cosine component:
0.50 m/s * cos(50°) = 0.321 m/s
Then we can apply the conservation of momentum in the horizontal direction:
(0.300 kg)(3.00 m/s) + (0.300 kg)(0) = (0.300 kg)(0.321 m/s) + (0.300 kg)(v2_final_horizontal)
Solving this, we find the horizontal component of ball 2's final velocity:
v2_final_horizontal = 2.679 m/s
Since there's no vertical motion before the collision, the vertical momentum must be conserved separately, resulting in ball 2 having no vertical velocity component after the collision. Thus, the final velocity of ball 2 will only have a horizontal component, and we find that it is
2.679 m/s to the right.