To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.
a) After the collision, the velocities of balls 2 and 3 can be determined using the conservation of momentum. Since ball 2 and 3 are initially at rest, their total momentum is zero. Therefore, the total momentum before the collision is equal to the total momentum after the collision.
Let's denote the masses of all three balls as m.
The momentum of ball 1 before the collision is given by:
Momentum1 = mass1 * velocity1 = m * 10 m/s = 10m
After the collision, the momentum of ball 1 is reversed in direction.
The total momentum of balls 2 and 3 after the collision is given by:
Total momentum2+3 = mass2 * velocity2 + mass3 * velocity3
Since the balls are identical, their masses are equal (m).
Hence, we have:
Momentum1 = Total momentum2+3
10m = m * velocity2 + m * velocity3
10 = velocity2 + velocity3 --------- Equation (1)
Note that since the collision is elastic, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. The kinetic energy can be calculated using the formula:
Kinetic energy = (1/2) * mass * velocity^2
The initial kinetic energy before the collision is given by:
Initial Kinetic Energy1 = (1/2) * m * (10 m/s)^2 = 50m^2
The final kinetic energy after the collision is given by:
Final Kinetic Energy2+3 = (1/2) * m * velocity2^2 + (1/2) * m * velocity3^2
Since the balls are identical, their masses are equal (m).
Hence, we have:
Initial Kinetic Energy1 = Final Kinetic Energy2+3
50m^2 = (1/2) * m * velocity2^2 + (1/2) * m * velocity3^2
50 = (1/2) * velocity2^2 + (1/2) * velocity3^2
100 = velocity2^2 + velocity3^2 ------------ Equation (2)
Now, we have two equations (Equation 1 and Equation 2) with two unknowns (velocity2 and velocity3). Let's solve them simultaneously.
Substituting velocity3 from Equation 1 into Equation 2:
100 = (velocity2 + 10)^2 + velocity2^2
100 = velocity2^2 + 20 velocity2 + 100 + velocity2^2
200 = 2 velocity2^2 + 20 velocity2
2 velocity2^2 + 20 velocity2 - 200 = 0
Dividing by 2:
velocity2^2 + 10 velocity2 - 100 = 0
Solving this quadratic equation, we find two solutions for velocity2:
velocity2 = 5 m/s or velocity2 = -20 m/s
Since velocity cannot be negative in this context, we discard the negative value.
Therefore, the velocity of ball 2 after the collision is 5 m/s.
Using Equation 1, we can now find the velocity of ball 3:
10 = 5 + velocity3
velocity3 = 5 m/s
Therefore, the velocities of balls 2 and 3 after the collision are both 5 m/s.
b) The velocity of ball 1 after the collision can be determined by using the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
The initial momentum of ball 1 is given by:
Momentum1 = mass1 * velocity1 = m * 10 m/s = 10m
The final momentum of ball 1 is given by:
Final Momentum1 = mass1 * final velocity1
Since the momentum is conserved, we have:
Momentum1 = Final Momentum1
10m = m * final velocity1
final velocity1 = 10 m/s
Therefore, the velocity of ball 1 after the collision is 10 m/s.