Question

A bullet of mass 20 gram traveling hoirzentally at 100 m/s embeds itself in the center of a block of wood mass 1 kg which is suspended by light vertical string 1 meter in length get cleared to maximum inclination of the string to the vertical​?

Answer 1

Answer:

36.5°

Explanation:

Momentum is conserved immediately before and after the collision.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(0.020 kg) (100 m/s) + (1 kg) (0 m/s) = (0.020 kg) v + (1 kg) v

v = 1.96 m/s

After the collision, energy is conserved.

KE = PE

½ m v² = mgh

½ v² = gh

h = v² / (2g)

h = (1.96 m/s)² / (2 × 9.8 m/s²)

h = 0.196 m

We need the angle that the string forms with the vertical.  Knowing the length of the string and the height that the block rises, we can do some geometry (see attached diagram).

L − L cos θ = h

L (1 − cos θ) = h

1 − cos θ = h / L

cos θ = 1 − h / L

cos θ = 1 − (0.196 m / 1 m)

θ = 36.5°

Round as needed.