An 80 g, 40 cm long rod hangs vertically on a frictionless, horizontal axle passing through its center. a 15 g ball of clay traveling horizontally at 2.0 m/s hits and sticks to the very bottom tip of the rod. to what maximum angle, measured from vertical, does the rod (with the attached ball of clay) rotate?

θ  = 12.95º

Explanation:

For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height

Let's start by finding the speed of the bar plus clay ball system, using amount of momentum

The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)

Initial before the crash

      p₀ = m v₀

Final after the crash before starting the movement

      = (m + M) v

     p₀ =

     m v₀ = (m + M) v

     v = v₀ m / (m + M)

     v = 2.0 0.015 / (0.015 +0.080)

     v = 0.316 m / s

With this speed the clay plus bar system comes out, let's use the concept of conservation of mechanical energy

Lower

    Em₀ = K = ½ (m + M) v²

Higher

    = U = (m + M) g y

   Em₀ =

   ½ (m + M) v² = (m + M) g y

   y = ½ v² / g

   y = ½ 0.316² / 9.8

   y = 0.00509 m

Let's look for the angle the height from the pivot point is

    L = 0.40 / 2 = 0.20 cm

The distance that went up is

     y = L - L cos θ

     cos θ  = (L-y) / L

     θ  = cos⁻¹ (L-y) / L

     θ  = cos⁻¹-1 ((0.20 - 0.00509) /0.20)

      θ  = 12.95º