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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Physics
Chapter
Laws of Motion

BITSAT 2019 · Physics · Laws of Motion

A body of mass m is moving in a straight line with momentum p ....

A body of mass mm is moving in a straight line with momentum pp. Starting at time t=0t = 0, a force F=atF = at acts in the same direction on the moving particle during time interval of TT. So that its momentum changes from pp to 2p2p. The value of TT is

Options

  1. A2pa\sqrt{\dfrac{2p}{a}}
  2. Bpa\sqrt{\dfrac{p}{a}}
  3. C22pa2\sqrt{\dfrac{2p}{a}}
  4. D2pa\dfrac{2p}{a}

Correct answer

A. 2pa\sqrt{\dfrac{2p}{a}}

Step-by-step solution

Given, F=atF = at

when, t=0t = 0, then linear momentum =p= p

when, t=Tt = T,~then linear momentum =2p= 2p

According to Newton's law of motion, applied force, F=dpdtF = \dfrac{dp}{dt}

or~dp=Fdtdp = Fdt

dp=atdtdp = atdt

or~∫p2pdp=a∫0Ttdt{∫}_p^{2p}dp = a{∫}_{0}^{T}tdt

\left(2p - p\right) = a \cdot \dfrac{T2}{2}</p><p></p><p>⇒ \dfrac{2p}{a} = T^{2} ⇒ T = \sqrt{\dfrac{2p}{a}}$

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