Step-by-step explanation:
To solve this problem, we can use the principle of conservation of momentum for an elastic collision. In an elastic collision, both momentum and kinetic energy are conserved.
The momentum of an object is given by the product of its mass and velocity: momentum = mass × velocity.
Before the collision, the total momentum of the system is the sum of the momenta of the basketball and the tennis ball:
Total momentum before collision = (mass of basketball × velocity of basketball) + (mass of tennis ball × velocity of tennis ball)
After the collision, the total momentum remains the same. Let's denote the velocity of the basketball after the collision as "v" (to the right).
Total momentum after collision = (mass of basketball × velocity of basketball after collision) + (mass of tennis ball × velocity of tennis ball after collision)
Using the conservation of momentum principle, we can set the total momentum before collision equal to the total momentum after collision:
(mass of basketball × initial velocity of basketball) + (mass of tennis ball × initial velocity of tennis ball) = (mass of basketball × velocity of basketball after collision) + (mass of tennis ball × velocity of tennis ball after collision)
Now, let's plug in the given values:
Mass of basketball (m1) = 0.6 kg
Initial velocity of basketball (u1) = 7.2 m/s
Mass of tennis ball (m2) = 0.04 kg
Initial velocity of tennis ball (u2) = -22 m/s (negative because it's moving to the left)
Velocity of basketball after collision (v) = unknown
Total momentum before collision = (0.6 kg × 7.2 m/s) + (0.04 kg × (-22 m/s)) = 4.32 kg m/s - 0.88 kg m/s = 3.44 kg m/s (to the right)
Total momentum after collision = (0.6 kg × v) + (0.04 kg × 32.7 m/s) = 0.6v + 1.308 kg m/s (to the right)
Now, set the total momentum before collision equal to the total momentum after collision:
3.44 kg m/s = 0.6v + 1.308 kg m/s
Subtract 1.308 kg m/s from both sides:
3.44 kg m/s - 1.308 kg m/s = 0.6v
2.132 kg m/s = 0.6v
Now, divide by 0.6:
v = 2.132 kg m/s / 0.6
v ≈ 3.553 m/s
So, the velocity of the basketball after the collision is approximately 3.553 m/s to the right.