Question

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Answer 1

Question:

A bullet of mass 20 g and moving with 600 m/s collides with a block of mass 4 kg hanging with the string. What is the velocity of the bullet when it comes out of the block, if block rises to height 0.2 m after collission?

Answer:

$\Large{\underline{\underline{\bf{Option\:A:200\:m/s}}}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Mass of the bullet (m₁ ) = 20 g = 0.02 kg
• Velocity of bullet before collisssion (u₁) = 600 m/s
• Mass of block (m₂) = 4 kg
• Height block rises to (h) = 0.2 m

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Velocity of bullet after collission (u₂)

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ First we need to find out the velocity of the wooden block(v).

→ Here the K.E possessed by the body is equal to P.E

  $\dfrac{1}{2} m_2v^{2} = m_2gh$

→ Cancelling mass on both sides

  $\dfrac{1}{2}v^{2} =gh$

Bringing 1/2 and square to the RHS

$v=\sqrt{2gh}$ where g = 9.8 m/s²

→ Substituting the given datas we get,

$v=\sqrt{2\times 9.8\times 0.2}$

v = 2 m/s

→ By the law of conservation of momentum, momentum before collission is equal to momentum after collission

m₁u₁ = m₁u₂ + m₂v

→ Substituting the datas we get,

0.02 × 600 = (0.02 × u₂) + (4 × 2)

12 = 0.02 u₂ +8

u₂ = 4/0.02

u₂ = 200 m/s

$\boxed{\bold{Velocity\:of\:bullet\:=200\:m/s}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ Momentum is always conserved in an isolated system. That is, the momentum before collission is equal to the momentum after collission.