Answer:
Step-by-step explanation:
In an absolutely elastic collision, both momentum and kinetic energy are conserved. We can use these conservation laws to solve for the final velocity of the second ball after the collision.
Let's start by calculating the initial momentum of the system. The momentum (p) of each ball is given by the product of its mass (m) and velocity (v):
p1 = m1v1 = 0.2 kg × 9 m/s = 1.8 kg m/s
p2 = m2v2 = 0.1 kg × 0 m/s = 0 kg m/s (the second ball is stationary)
The total initial momentum of the system is:
p1 + p2 = 1.8 kg m/s + 0 kg m/s = 1.8 kg m/s
After the collision, the first ball bounces back with a velocity of 4 m/s. Let's call the final velocity of the second ball v2. The total final momentum of the system is:
p1' + p2' = m1v1' + m2v2'
where v1' = -4 m/s (negative sign indicates the opposite direction) and v2' = v2 (since the second ball continues to move in the same direction).
The total final momentum is also equal to the initial momentum, since momentum is conserved:
p1' + p2' = p1 + p2
m1v1' + m2v2' = 1.8 kg m/s
Now we can use the conservation of kinetic energy to solve for v2. The total initial kinetic energy (K) of the system is:
K = 0.5 m1v1^2 + 0.5 m2v2^2
Since the collision is absolutely elastic, the total final kinetic energy is the same as the initial kinetic energy:
K' = 0.5 m1v1'^2 + 0.5 m2v2'^2 = K
Substituting the given values, we get:
0.5 × 0.2 kg × (9 m/s)^2 + 0.5 × 0.1 kg × 0 m/s^2 = 0.5 × 0.2 kg × (4 m/s)^2 + 0.5 × 0.1 kg × v2'^2
Simplifying, we get:
81 J = 8 J + 0.05 kg v2'^2
Solving for v2', we get:
v2' = √(73.8) m/s ≈ 8.6 m/s
Therefore, the final velocity of the second ball after the collision is approximately 8.6 m/s.