The velocity of both the person and the ball after the interaction is 2.0 m/s
The Breakdown
To solve this problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the interaction is equal to the total momentum after the interaction.
Let's denote the initial velocity of the medicine ball as v1 and the initial velocity of the person as v2. After the person catches the ball, they both move together with a common final velocity, which we'll denote as vf.
The momentum of the medicine ball before the interaction is given by:
p1 = m1 × v1
The momentum of the person before the interaction is given by:
p2 = m2 × v2
The total momentum before the interaction is:
p_initial = p1 + p2
Since momentum is conserved, the total momentum after the interaction is also equal to p_initial:
p_final = p_initial
After the person catches the ball, they both move together with a common final velocity vf. The total momentum after the interaction is given by:
p_final = (m1 + m2) × vf
Setting p_initial equal to p_final, we have:
p_initial = p_final
m1 × v1 + m2 × v2 = (m1 + m2) × vf
Now we can substitute the given values into the equation and solve for vf:
m1 = 15.0 kg (mass of the medicine ball)
v1 = 10.0 m/s (initial velocity of the medicine ball)
m2 = 60.0 kg (mass of the person)
v2 = 0 m/s (initial velocity of the person at rest)
(15.0 kg × 10.0 m/s) + (60.0 kg × 0 m/s) = (15.0 kg + 60.0 kg) × vf
150 kg·m/s = 75 kg vf
Dividing both sides by 75 kg:
vf = 150 kg·m/s / 75 kg
vf = 2.0 m/s
Therefore, the velocity of both the person and the ball after the interaction is 2.0 m/s.