Question

A bullet with mass 4.00 g and speed 650.0 m/s is fired at a block of wood with mass 0.0950 kg. The block rests on a "frictionless" surface and is thin enough so that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is 23.0 m/s. (a) What is the speed of the bullet after it exits the block? (answer: 104 m/s)

Answer 1

Answer:

The final speed of the bullet after exit = 104 m/s²

Explanation:

Law of conservation of momentum :

The law of conservation of momentum states that in a system of colliding object, the total momentum is conserved, provided there is no net external force acting on the system. I.e

Total momentum before collision = total momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ +m₂v₂............................equation 1

Making v₂ the subject of the equation,

v₂ ={(m₁u₁ + m₂u₂) - m₂v₂}/m₁ .......................equation 2

Where m₁  mass of the bullet, m₂ = mass of the block, u₁ = initial velocity of the bullet, u₂ = initial velocity of the block, v₁ = final velocity of the bullet, v₂ = final velocity of the block.

From the question,

m₁ = 4.00 g = 4/1000 = 0.004 kg, m₂ = 0.0950 kg, u₁ = 650 m/s, u₂ = 0 m/s  (the block was initially at rest before the bullet was fired at it), v₂ = 23 m/s

Substituting these values into equation 2,

v₁ = [{(0.004 × 650)+(0.0950 × 0)}- (0.0950 × 23)]/0.004

v₁ = (2.6 + 0 - 2.185)/0.004

v₁ = 0.415/0.004 = 103.75

v₁ ≈ 104 m/s

Answer 2

Answer:

26.4 m/s

Explanation:

4 g = 0.004 kg

By the law of momentum conservation, the momentum before and after the impact must be the same

$P_{before} = P_{after}$

$0.004*650 = 0.004*23 + 0.095*v_b$

$v_b = \frac{2.508}{0.095} = 26.4 m/s$

So the speed of the block after the impact is 26.4 m/s