A billiard ball (Ball #1) moving at 5.00 m/s strikes a stationary ball (Ball #2) of the same mass. After the elastic collision, Ball #1 moves at a speed of 3.50 m/s. Find the speed of Ball #2 after the collision.Select one:a.3.57 m/sb.2.16 m/sc.1.25 m/sd.1.50 m/s
Question
A billiard ball (Ball #1) moving at 5.00 m/s strikes a stationary ball (Ball #2) of the same mass. After the elastic collision, Ball #1 moves at a speed of 3.50 m/s. Find the speed of Ball #2 after the collision.Select one:a.3.57 m/sb.2.16 m/sc.1.25 m/sd.1.50 m/s
Solution
This problem can be solved using the principles of conservation of momentum and conservation of kinetic energy, which are both applicable in an elastic collision.
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Conservation of momentum: The total momentum before the collision is equal to the total momentum after the collision. If we let m1 be the mass of Ball #1, m2 be the mass of Ball #2, v1i be the initial velocity of Ball #1, v2i be the initial velocity of Ball #2, v1f be the final velocity of Ball #1, and v2f be the final velocity of Ball #2, then we can write the conservation of momentum as:
m1v1i + m2v2i = m1v1f + m2v2f
Given that the balls have the same mass and that Ball #2 is initially stationary, this simplifies to:
v1i = v1f + v2f
Substituting the given values:
5.00 m/s = 3.50 m/s + v2f
Solving for v2f gives:
v2f = 5.00 m/s - 3.50 m/s = 1.50 m/s
So, the speed of Ball #2 after the collision is 1.50 m/s. The correct answer is (d) 1.50 m/s.
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