two billiard balls of equal mass undergo a perfectly elastic collision. if the speed of one ball was initially 5.72 m/s and the other 1.64 m/s. what will their speeds be after the collision
To find the speeds of the billiard balls after the collision, we can use the principle of conservation of momentum and kinetic energy.
1. Conservation of Momentum:
When two objects collide, the total momentum before the collision is equal to the total momentum after the collision, provided no external forces act on the system.
Mathematically, we can express this as:
m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f
Where:
m1 and m2 are the masses of the first and second ball, respectively.
v1i and v2i are the initial velocities of the first and second ball, respectively.
v1f and v2f are the final velocities of the first and second ball, respectively.
2. Conservation of Kinetic Energy:
In a perfectly elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Mathematically, we can express this as:
0.5 * m1 * (v1i)^2 + 0.5 * m2 * (v2i)^2 = 0.5 * m1 * (v1f)^2 + 0.5 * m2 * (v2f)^2
Now, let's solve the above equations to find the final velocities:
Given:
m1 = m2 (equal mass)
v1i = 5.72 m/s
v2i = 1.64 m/s
Using the conservation of momentum equation:
m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f
Since m1 = m2, we can simplify the equation to:
v1i + v2i = v1f + v2f
Substituting the given values:
5.72 m/s + 1.64 m/s = v1f + v2f
7.36 m/s = v1f + v2f
Now, using the conservation of kinetic energy equation:
0.5 * m1 * (v1i)^2 + 0.5 * m2 * (v2i)^2 = 0.5 * m1 * (v1f)^2 + 0.5 * m2 * (v2f)^2
Substituting the given values:
0.5 * v1i^2 + 0.5 * v2i^2 = 0.5 * v1f^2 + 0.5 * v2f^2
0.5 * (5.72 m/s)^2 + 0.5 * (1.64 m/s)^2 = 0.5 * v1f^2 + 0.5 * v2f^2
0.5 * (32.7984 m/s^2) + 0.5 * (2.6896 m/s^2) = 0.5 * v1f^2 + 0.5 * v2f^2
16.3992 m/s^2 + 1.3448 m/s^2 = 0.5 * v1f^2 + 0.5 * v2f^2
17.744 m/s^2 = 0.5 * v1f^2 + 0.5 * v2f^2
Substituting v1f + v2f = 7.36 m/s:
17.744 m/s^2 = 0.5 * (7.36 m/s - v2f)^2 + 0.5 * v2f^2
Expanding and simplifying the equation:
17.744 m/s^2 = 0.5 * (54.1696 m/s^2 - 14.3488 m/s * v2f + v2f^2) + 0.5 * v2f^2
17.744 m/s^2 = 27.0848 m/s^2 - 7.1744 m/s * v2f + 0.5 * v2f^2 + 0.5 * v2f^2
17.744 m/s^2 = 27.0848 m/s^2 - 7.1744 m/s * v2f + v2f^2
Rearranging the equation:
2v2f^2 - 7.1744 m/s * v2f + (27.0848 m/s^2 - 17.744 m/s^2) = 0
2v2f^2 - 7.1744 m/s * v2f + 9.3408 m/s^2 = 0
Now we can solve this quadratic equation to find the values of v2f (final velocity of the second ball) using the quadratic formula:
v2f = [-(-7.1744 m/s) ± √((-7.1744 m/s)^2 - 4 * 2 * 9.3408 m/s^2)] / (2 * 2)
Simplifying and solving the equation, we find two possible values for v2f.
v2f ≈ 2.8418 m/s or v2f ≈ 1.3326 m/s
Since the collision is perfectly elastic, the final velocities of both balls will be the same. Therefore, the final velocities of both balls after the collision will be:
v1f = v2f ≈ 2.8418 m/s, or v1f = v2f ≈ 1.3326 m/s
So, the final speeds of the two balls after the collision will be approximately 2.8418 m/s and 1.3326 m/s.