In a circus an elderly performer thrills the crowd by catching a canon ball shot at him. The cannon ball has a mass of 28.0 kg and its horizontal component of velocity is 6.50 m/s when the 65.0 kg performer catches it. If the performer is on nearly frictionless roller skates, what is his recoil velocity?

To find the recoil velocity of the performer, we can use the principle of conservation of momentum. According to this principle, the total momentum before and after an event remains the same, as long as no external forces are acting on the system.

Let's assume the positive direction as the direction in which the performer catches the cannonball. Initially, the cannonball has a mass of 28.0 kg and a horizontal velocity of 6.50 m/s. The performer has a mass of 65.0 kg and an initial velocity of 0 m/s.

The conservation of momentum equation can be written as:

(mass of cannonball × initial velocity of cannonball) + (mass of performer × initial velocity of performer) = (mass of cannonball × final velocity of cannonball) + (mass of performer × final velocity of performer)

Plugging in the values:

(28.0 kg × 6.50 m/s) + (65.0 kg × 0 m/s) = (28.0 kg × final velocity of cannonball) + (65.0 kg × final velocity of performer)

182.0 kg·m/s = 28.0 kg × final velocity of cannonball + 65.0 kg × final velocity of performer

Since the performer catches the cannonball, the final velocity of the cannonball is zero. This means the term (28.0 kg × final velocity of the cannonball) becomes zero. Rearranging the equation, we get:

(65.0 kg × final velocity of performer) = 182.0 kg·m/s

Now, we can solve for the final velocity of the performer:

final velocity of performer = 182.0 kg·m/s / 65.0 kg

final velocity of performer = 2.80 m/s (rounded to two decimal places)

Therefore, the recoil velocity of the performer is approximately 2.80 m/s in the opposite direction to the cannonball's initial velocity.