During a circus act, a performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of m1 = 10.5 kg and the horizontal component of its velocity is v = 8.85 m/s when the performer, of mass m2 = 61.5 kg, catches it. If the performer is on nearly frictionless roller skates, what is his recoil velocity?

1.512 m/s

Explanation:

M₁ = 10.5

M₂ = 61.5kg

v₁ = 8.85m/s

v₂ = ?

This question involves conservation of momentum and law of conservation of momentum states that, the total momentum of a system does not change before and after impact and it's always constant.

Momentum = mass * velocity = M*V

total momentum of the system before caught the ball = total momentum after he caught the ball.

M₁V₁ = M₂V₂

10.5 * 8.85 = 61.5 * V₂

V₂ = 92.925 / 61.5

V₂ = 1.5109m/s ≈ 1.512m/s

9264706 molybdenum atoms have to be laid.

Explanation:

The parameter of the atom which  will determine the number of atoms needed to span a certain distance is the diameter. Diameter (d) is simply two times radius(r):

(eq. 1)

If we express the number of molybdenum atoms needed with the letter x, the spanned distance (2.52mm) has to be equal to the number of atoms (x) times the diameter of these atoms.

(eq 2.)

Before stating an equation, all units have to be consistent (272 picometres expressed in mm, or 2.52mm expressed in pm, either one). Let's convert 272pm, using a conversion factor:

Conversion factor:

Now, solving for "x" in the second equation:

Plugging in the "d" value: