In a perfectly elastic collision, both momentum and kinetic energy are conserved. Let's denote the initial velocity of the first ball (moving east) as "v1" and the initial velocity of the second ball (at rest) as "v2".
Since the second ball is initially at rest, its initial velocity, v2, is 0 m/s.
The conservation of momentum can be expressed as:
m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final
where: m1 = mass of the first ball = 1 kg (assuming the mass of the first ball is 1 kg) m2 = mass of the second ball = 2 kg
The conservation of kinetic energy can be expressed as:
(1/2) * m1 * v1_initial^2 + (1/2) * m2 * v2_initial^2 = (1/2) * m1 * v1_final^2 + (1/2) * m2 * v2_final^2
Plugging in the given values:
(1/2) * 1 kg * (5 m/s)^2 + (1/2) * 2 kg * (0 m/s)^2 = (1/2) * 1 kg * v1_final^2 + (1/2) * 2 kg * v2_final^2
Simplifying:
12.5 = (1/2) * 1 kg * v1_final^2 + 0
12.5 = (1/2) * 1 kg * v1_final^2
Dividing both sides by 1/2 kg:
25 = v1_final^2
Taking the square root of both sides:
v1_final = ±5 m/s
Since the collision is elastic, the first ball's velocity after the collision is 5 m/s in the opposite direction of its initial velocity. Therefore, v1_final = -5 m/s (moving west).
To find the velocity of the second ball (v2_final), we can use the equation for momentum conservation:
m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final
Plugging in the known values:
1 kg * 5 m/s + 2 kg * 0 m/s = 1 kg * (-5 m/s) + 2 kg * v2_final
5 kg m/s = -5 kg m/s + 2 kg * v2_final
10 kg m/s = 2 kg * v2_final
Dividing both sides by 2 kg:
5 m/s = v2_final
Therefore, the final velocity of the second ball (v2_final) is 5 m/s (moving east).
In summary, after the perfectly elastic collision:
- The first ball's velocity (v1_final) is -5 m/s (moving west).
- The second ball's velocity (v2_final) is 5 m/s (moving east).