Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

74 of 130

Question 74 of 130

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:

Show answer and solution

Answer: (A) 6Ns6Ns

The formula to calculate the impulse (J)\left(J\right) on the gun is given by
J=m(v−u)...(1)J=m\left(v-u\right)...\left(1\right)
where, mm is the mass, uu is the initial velocity and vv is the final velocity of the bullet.
Substitute the values of the known parameters into equation (1) to calculate the required impulse.
J=10g×1kg1000g×(600−0)ms−1=6Ns\begin{matrix}J & = & 10g\times \dfrac{1kg}{1000g}\times \left(600-0\right)m{s}^{-1} \\ & = & 6Ns\end{matrix}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
Revise Rotational Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Center of Mass, Momentum & Collision (Impulse), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.