Question

A 22-g bullet traveling 240 m/s penetrates a 2.0-kg block of wood and emerges going 150 m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

Answer 1

Answer:

After the bullet emerges the block moves at 0.99 m/s

Explanation:

Given;

mass of bullet, m₁ = 22 g = 0.022 kg

initial speed of the bullet, u₁ = 240 m/s

final speed of the bullet, v₁ = 150 m/s

mass of block, m₂ = 2.0 kg

initial speed of the block, u₂ = 0

Let the final speed of the block = v₂

Apply principles of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

0.022 x 240 + 2 x 0 = 0.022 x 150 + 2v₂

5.28 = 3.3 + 2v₂

5.28 - 3.3 = 2v₂

1.98 = 2v₂

v₂ = 1.98 / 2

v₂ = 0.99 m/s

Therefore, after the bullet emerges the block moves at 0.99 m/s

Answer 2

The body moves at a speed of 2.61m/s after the bullet emerges.

According to the law of collision which states that the momentum of the body before the collision is equal to the momentum of the body after the collision.

The formula for calculating the collision of a body is expressed as:

p = mv

m is the mass of the body

v is the velocity of the body

Based on the law above;

$m_1u_1+m_2u_2=(m_1+m_2)v$

v is the final velocity of the body after the collision

Substitute the given parameters into the formula as shown:

$0.022(240) + 2(0) = (0.022+2)v\\ 5.28 = 2.022v\\v=\frac{5.28}{2.022}\\v= 2.611m/s$

This shows that the body moves at a speed of 2.61m/s after the bullet emerges.

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