A 0.600-kg ball traveling 4.00 m/s to the right collides with a 1.00-kg ball traveling
5.00 m/s to the left. After the collision, the lighter ball is traveling 7.25 m/s to the
left. What is the velocity of the heavier ball after the collision?
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum of a system remains constant before and after a collision, provided no external forces act on the system during the collision.
The momentum of an object is defined as the product of its mass and velocity. Therefore, we can write the equation for conservation of momentum as:
(m1 * v1) + (m2 * v2) = (m1 * v1') + (m2 * v2')
Where m1 and m2 are the masses of the balls, v1 and v2 are their initial velocities, and v1' and v2' are their final velocities.
Given:
m1 = 0.600 kg (mass of the lighter ball)
v1 = 4.00 m/s (initial velocity of the lighter ball)
m2 = 1.00 kg (mass of the heavier ball)
v2 = -5.00 m/s (initial velocity of the heavier ball, negative because it's traveling to the left)
v1' = 0 m/s (final velocity of the lighter ball, since it comes to rest)
v2' = ? (final velocity of the heavier ball)
Substituting these values into the conservation of momentum equation:
(0.600 kg * 4.00 m/s) + (1.00 kg * (-5.00 m/s)) = (0.600 kg * 0 m/s) + (1.00 kg * v2')
Simplifying the equation:
2.40 kg m/s - 5.00 kg m/s = 0 kg m/s + 1.00 kg v2'
-2.60 kg m/s = 1.00 kg v2'
Now solve for v2':
v2' = (-2.60 kg m/s) / (1.00 kg)
v2' = -2.60 m/s
Therefore, the velocity of the heavier ball after the collision is -2.60 m/s to the left.