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The Linear momentum of a particle is p = ct2 + bt + a. Find the force on the particle at time t = 1 second. (Here a, b and c are constants)
1. 2c + a
2. b + 2c
3. 2a + b
4. a + b + 2c

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Best answer
Correct Answer - Option 2 : b + 2c

CONCEPT:

  • Newton’s Second Law of Motion: The rate of change of linear momentum of a body is directly proportional to the external force applied on the body and this change takes place always in the direction of the applied force.
  • According to the second law of motion, force

\(Force \propto \frac{{change\;in\;momentum}}{{time}}\)

\(⇒ F = K\frac{{{p_2} - {p_1}}}{t} = K\frac{{m\left( {v - u} \right)}}{t} = Kma\)          \(\left[ {\therefore \frac{{v - u}}{t} = a} \right]\)

Where, K = constant of proportionality, p = momentum of the body, m = mass of the body, u = initial velocity of the body, v = final velocity of the body, and a = acceleration of the body.

EXPLANATION:

Given that:

momentum (p) = ct2 + bt + a

  • According to the second law of motion, force

\(⇒ F =\frac{dp}{dt}\)

\(⇒ F = \frac{d(ct^2 \; + \; bt \; +\; a)}{dt}\)

⇒ F = 2ct + b

The force on the particle at t = 1 sec is 

F = 2c × 1 + b = b + 2c

  • Therefore option 2 is correct. 

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