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Karnataka Board PUCPUC Science Class 11

The motion of a particle of mass m is given by x = 0 for t < 0 s, x(t) = A sin 4 pt for 0 < t < (1/4) s (A > o), and x = 0 for t > (1/4) s. Which of the following statements is true?

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Question

The motion of a particle of mass m is given by x = 0 for t < 0 s, x(t) = A sin 4 pt for 0 < t < (1/4) s (A > o), and x = 0 for t > (1/4) s. Which of the following statements is true?

  1. The force at t = (1/8) s on the particle is – 16π2 Am.
  2. The particle is acted upon by on impulse of magnitude 4π2 A m at t = 0 s and t = (1/4) s.
  3. The particle is not acted upon by any force.
  4. The particle is not acted upon by a constant force.
  5. There is no impulse acting on the particle.
Short/Brief Note
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Solution

a, b and d

Explanation:

Given, `x = 0` for `t < 0  s`.

`x(t) = A sin 4pit`; for `0 < t < 1/4  s`

`x = 0`; for `t > 1/4  s`

For, `0 < t < 1/4  s`

`v(t) = (dx)/(dt) = 4π A cos 4πt`

`a(t)` = acceleration

= `(dv(t))/(dt)`

= – 16π2 A sin 4 πt

At `t = 1/8  s, a(t) = - 16 π^2  A sin 4π xx 1/8 = - 16π^2  A`

`F = ma (t) = - 16 π^2` A × m = `- 16 π^2` mA

Impulse = Change in linear momentum

= `F xx t = (- 16 π^2  Am) xx 1/4`

= `- 4π^2  Am`

The impulse (Change in linear momentum) at t = 0 is the same as, t = `1/4  s`.

Clearly, force depends upon A which is not constant. Hence, force is also not constant.

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Chapter 5: Laws of Motion - Exercises [Page 31]

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NCERT Exemplar Physics [English] Class 11
Chapter 5 Laws of Motion
Exercises | Q 5.10 | Page 31

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