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JEE Main 202313 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
13 April
Shift
Morning Shift
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2023 · 13 April · Morning Shift · Physics · Center of Mass, Momentum and Collision

A bullet of 10g leaves the barrel of gun with a velocity of 600m s...

A bullet of 10g10g leaves the barrel of gun with a velocity of 600ms−1600m{s}^{- 1}. If the barrel of gun is 50cm50cm long and mass of gun is 3kg,3kg, then value of impulse supplied to the gun will be:

Options

  1. A6Ns6Ns
  2. B3Ns3Ns
  3. C36Ns36Ns
  4. D12Ns12Ns

Correct answer

A. 6Ns6Ns

Step-by-step solution

The formula to calculate the impulse (J)\left(J\right) on the gun is given by

J=m(v−u)...(1)</p><p>where,J = m\left(v - u\right) . . . \left(1\right)</p><p>where,misthemass,is the mass,uistheinitialvelocityandis the initial velocity andvisthefinalvelocityofthebullet.</p><p>Substitutethevaluesoftheknownparametersintoequation(1)tocalculatetherequiredimpulse.</p><p>is the final velocity of the bullet.</p><p>Substitute the values of the known parameters into equation (1) to calculate the required impulse.</p><p>\begin{cases}J & = & 10g \times \dfrac{1kg}{1000g} \times \left(600 - 0\right)m{s}^{- 1} \\ & = & 6Ns\end{cases}</p></blockquote></p></blockquote>

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