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Center of Mass question

2022 · 26 Jun · Shift 2 · Q65
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Center of Mass question

2022 · 26 Jun · Shift 2 · Q65

JEE MainPhysicsCenter of MassNumerical+4 / −1
A batsman hits back a ball of mass 0.4 kg straight in the direction of the bowler without changing its initial speed of 15 ms −-− 1. The impulse imparted to the ball is ‾\underline{\hspace{2cm}}​ Ns.
Numerical answer
View written solutionFree

Correct answer: 12

  1. Understand the motion

The ball is initially coming towards the batsman with speed 15 m s−115\,\text{m s}^{-1}15m s−1. After being hit, it goes back towards the bowler with the same speed 15 m s−115\,\text{m s}^{-1}15m s−1.

So the velocity reverses direction.

  1. Choose a sign convention

Take the direction towards the bowler as positive.

Then, u=−15 m s−1,v=+15 m s−1u = -15\,\text{m s}^{-1}, \qquad v = +15\,\text{m s}^{-1}u=−15m s−1,v=+15m s−1

  1. Use impulse-momentum theorem

Impulse is equal to change in momentum: J=Δp=m(v−u)J = \Delta p = m(v-u)J=Δp=m(v−u)

Given mass of ball, m=0.4 kgm = 0.4\,\text{kg}m=0.4kg

Therefore, J=0.4 [15−(−15)]J = 0.4\,[15-(-15)]J=0.4[15−(−15)] J=0.4×30J = 0.4\times 30J=0.4×30 J=12 N sJ = 12\,\text{N s}J=12N s

  1. Final answer

The impulse imparted to the ball is 12 N s\boxed{12\,\text{N s}}12N s​

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