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Center of Mass question

2019 · 11 Jan · Shift 2 · Q57
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Center of Mass question

2019 · 11 Jan · Shift 2 · Q57

JEE MainPhysicsCenter of MassMCQ+4 / −1
A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here k is a constant. The value of T is :
  1. A
    2kp2\sqrt {{k \over p}}2pk​​
  2. B
    2pk2\sqrt {{p \over k}}2kp​​
  3. C
    2p2\sqrt {{{2p} \over 2}}22p​​
  4. D
    2kp\sqrt {{{2k} \over p}}p2k​​
View written solutionFree

Correct answer: B

  1. Given data
  • Initial momentum: ppp
  • Final momentum: 3p3p3p
  • Force applied: F=ktF = ktF=kt
  • Force acts from t=0t=0t=0 to t=Tt=Tt=T

We use the relation between force and momentum:

F=dpdtF = \frac{dp}{dt}F=dtdp​

So,

dpdt=kt\frac{dp}{dt} = ktdtdp​=kt

  1. Integrate over the time interval

The momentum changes from ppp to 3p3p3p in time 000 to TTT:

∫p3pdp=∫0Tkt dt\int_p^{3p} dp = \int_0^T kt\,dt∫p3p​dp=∫0T​ktdt

Left side:

∫p3pdp=3p−p=2p\int_p^{3p} dp = 3p - p = 2p∫p3p​dp=3p−p=2p

Right side:

∫0Tkt dt=k∫0Tt dt=k[t22]0T=kT22\int_0^T kt\,dt = k\int_0^T t\,dt = k\left[\frac{t^2}{2}\right]_0^T = \frac{kT^2}{2}∫0T​ktdt=k∫0T​tdt=k[2t2​]0T​=2kT2​

Thus,

2p=kT222p = \frac{kT^2}{2}2p=2kT2​

  1. Solve for TTT

kT2=4pkT^2 = 4pkT2=4p

T2=4pkT^2 = \frac{4p}{k}T2=k4p​

T=2pkT = 2\sqrt{\frac{p}{k}}T=2kp​​

  1. Match with options

This matches Option B:

2pk2\sqrt{\frac{p}{k}}2kp​​

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.

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