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An 8kg bullet is shot into a 4kg block , at rest on a frictionless horizontal surface . The bullet remains lodged in the block.The block moves into an ideal massless spring and compresses it by 8, 7cm. The spring constant of the spring is 2400n/m.What is the initial speed of the bullet?

Sagot :

Lanuel

By applying the law of conservation of energy, the initial speed of this bullet is equal to 1.8456 m/s.

Given the following data:

  • Mass of the bullet, m₁ = 8 kg
  • Mass of the block, m₂ = 4 kg.
  • Initial speed of block = 0 m/s (since it's at rest).
  • Compression in the spring, x = 8.7 cm to m = 8.7 × 10 m
  • Spring constant, k = 2400 N/m.

How to calculate initial speed of the bullet?

First of all, we would apply the law of conservation of momentum to derive an expression for the final velocity of both objects (bullet and block):

m₁u₁ + m₂u₂ = (m₁ + m₂)v

m₁u₁ + 0 = (m₁ + m₂)v

m₁u₁ = (m₁ + m₂)v

v = m₁u₁/(m₁ + m₂)

In order to calculate the initial speed of this bullet, we would also apply the law of conservation of energy:

K.E = U

1/2 × (m₁ + m₂)v² = 1/2kx²

Substituting the value of v, we have:

(m₁ + m₂)[m₁u₁/(m₁ + m₂)]² = kx²

m₁²u₁²/(m₁ + m₂) = kx²

m₁²u₁² = (m₁ + m₂)kx²

u₁² = [(m₁ + m₂)kx²]/m₁²

u₁ = √([(m₁ + m₂)kx²]/m₁²)

u₁ = √([(8 + 4) × 2400 × 0.087²]/8²)

u₁ = √([12 × 2400 × 0.007569]/64)

u₁ = √217.9872/64

u₁ = √3.4061

Initial speed, u₁ = 1.8456 m/s.

Read more on initial speed here: https://brainly.com/question/13966700

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