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
[tex]P_{before} = P_{after}[/tex]
[tex]0.004*650 = 0.004*23 + 0.095*v_b[/tex]
[tex]v_b = \frac{2.508}{0.095} = 26.4 m/s[/tex]
So the speed of the block after the impact is 26.4 m/s
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
