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In the ballistic pendulum shown below, the block has a mass of 1.0 kg, the bullet has a mass of 0.0052 kg, and the string has a length of 1.57 m. A bullet is shot directly at the block, lodging into the block, and causing the pendulum to swing forward 26°.
What was the initial speed of the bullet, in m/s?

In the ballistic pendulum shown below the block has a mass of 10 kg the bullet has a mass of 00052 kg and the string has a length of 157 m A bullet is shot dire class=


Answer :

MathPhys

Answer:

341 m/s

Explanation:

During the collision, the momentum of the system is conserved. After the collision, the system's initial kinetic energy is converted to gravitational potential energy.

Conservation of energy:

PE = KE

mgh = ½ mv²

v² = 2gh

v² = 2g (L − L cos θ)

Plug in values and solve.

v² = 2 (9.8) (1.57 − 1.57 cos 26°)

v = 1.765 m/s

Conservation of momentum:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

The block's initial speed is zero, so u₂ = 0. After the collision, the bullet and block have the same speed, so v₁ = v₂ = v.

m₁u₁ = (m₁ + m₂) v

Plug in values and solve:

(0.0052) u₁ = (0.0052 + 1.0) (1.765)

u₁ = 341 m/s

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