To determine the velocities of the three balls after the collision, we can apply the law of conservation of momentum and the law of conservation of kinetic energy. Here's how we can approach it step by step:
1. Assign variables:
Let V be the initial speed of ball 1 and v1, v2, and v3 be the velocities of balls 1, 2, and 3 after the collision, respectively.
2. Law of Conservation of Momentum:
In an elastic collision, the total momentum before and after the collision remains conserved. The initial momentum is given by:
Initial momentum = mass of ball 1 * velocity of ball 1
(Since all three balls are identical, their masses are the same.)
Therefore, the initial momentum is m * V, where m is the mass of any ball.
After the collision, the final momentum is given by:
Final momentum = mass of ball 1 * velocity of ball 1 + mass of ball 2 * velocity of ball 2 + mass of ball 3 * velocity of ball 3
Since the balls are in contact and moving in a line perpendicular to the initial velocity of ball 1, the momentum of ball 2 and ball 3 together is zero, as they move together.
Therefore, the final momentum simplifies to:
Final momentum = mass of ball 1 * velocity of ball 1 + 0 + 0
= mass of ball 1 * velocity of ball 1
= m * v1 (since v2 = v3 = 0)
Setting the initial and final momenta equal, we get:
m * V = m * v1
Canceling out the mass, we obtain:
V = v1 ... Equation 1
3. Law of Conservation of Kinetic Energy:
In an elastic collision, the total kinetic energy before and after the collision remains conserved.
The initial kinetic energy is given by:
Initial kinetic energy = (1/2) * mass of ball 1 * (velocity of ball 1)^2
= (1/2) * m * V^2
After the collision, the final kinetic energy is given by:
Final kinetic energy = (1/2) * mass of ball 1 * (velocity of ball 1)^2 + (1/2) * mass of ball 2 * (velocity of ball 2)^2 + (1/2) * mass of ball 3 * (velocity of ball 3)^2
Since ball 2 and ball 3 move together, their velocities have the same magnitude but in opposite directions.
Therefore, the final kinetic energy simplifies to:
Final kinetic energy = (1/2) * mass of ball 1 * (velocity of ball 1)^2 + (1/2) * mass of balls 2 & 3 * (velocity of ball 2)^2
Substituting the values, we have:
Final kinetic energy = (1/2) * m * (v1)^2 + (1/2) * 2m * (v2)^2 (since mass of ball 2 = mass of ball 3 = m)
Simplifying further, we get:
Final kinetic energy = (1/2) * m * (v1)^2 + m * (v2)^2
Since the balls are frictionless, the kinetic energy remains conserved.
Therefore, the initial and final kinetic energies are equal, giving us:
(1/2) * m * V^2 = (1/2) * m * (v1)^2 + m * (v2)^2
Cancelling out the mass and rearranging, we have:
V^2 = (v1)^2 + 2 * (v2)^2 ... Equation 2
4. Solving the equations:
We have Equation 1: V = v1
And Equation 2: V^2 = (v1)^2 + 2 * (v2)^2
Substituting V = v1 in Equation 2, we get:
(v1)^2 = (v1)^2 + 2 * (v2)^2
Simplifying, we find:
0 = 2 * (v2)^2
Since (v2)^2 cannot be negative, the only solution is v2 = 0.
Using v2 = 0 in Equation 1, we have:
V = v1 => v1 = V
Therefore, the velocities of the three balls after the collision are:
v1 = V
v2 = 0
v3 = 0