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Chapter 3: Problem 366

The linear momentum \(\mathrm{P}\) of a particle varies with the time as follows. \(P=a+b t^{2}\) Where \(a\) and \(b\) are constants. The net force acting on the particle is (A) Proportional to t (B) Proportional to t \(^{2}\) (C) Zero (D) constant

Short Answer

Expert verified
The net force (F) acting on the particle is proportional to time (t), as found by differentiating the given momentum expression: \(F = \frac{dP}{dt} = 2bt\). Therefore, the correct answer is (A) Proportional to t.

Step by step solution

01

1) Differentiate the Momentum Expression

First, we need to find the time derivative of the momentum expression P = a + b*t^2. To do this, we will apply the power rule of differentiation for each term concerning time: \(\frac{dP}{dt} = \frac{d}{dt}(a + bt^2)\)
02

2) Differentiate the Constants

Since 'a' and 'b' are constants, the derivative of the first term with respect to time is zero: \(\frac{d}{dt}(a) = 0\)
03

3) Differentiate the Second Term

Now, we need to differentiate the second term concerning time: \(\frac{d}{dt}(bt^2)= b\frac{d}{dt}(t^2) = 2bt\)
04

4) Combine the Results

Now, let's combine our results from steps 2 and 3: \(\frac{dP}{dt}= 0 + 2bt = 2bt\)
05

5) Identify the Net Force

We know that the net force acting on the particle (F) is equal to the time derivative of momentum: \(F =\frac{dP}{dt}\) So, substituting our result for the derivative: \(F = 2bt\)
06

6) Final Analysis

Now, we can see that the net force (F) acting on the particle is proportional to time (t) since it is directly proportional to 't' multiplied by a constant '2b'. Consequently, the correct option is: (A) Proportional to t

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Most popular questions from this chapter

Force of \(5 \mathrm{~N}\) acts on a body of weight \(9.8 \mathrm{~N}\). what is the acceleration produced in \(\mathrm{ms}^{-2}\). (A) \(49.00\) (B) \(5.00\) (C) \(1.46\) (D) \(0.51\)

A wagon weighing \(1000 \mathrm{~kg}\) is moving with a velocity $50 \mathrm{~km} \mathrm{~h}^{-1}\( on smooth horizontal rails. A mass of \)250 \mathrm{~kg}$ is dropped into it. The velocity with which it moves now is (A) \(2.5 \mathrm{~km} \mathrm{~h}^{-1}\) (B) \(20 \mathrm{~km} \mathrm{~h}^{-1}\) (C) \(40 \mathrm{~km} \mathrm{~h}^{-1}\) (D) \(50 \mathrm{~km} \mathrm{~h}^{-1}\)

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A Particle moves in the X-Y plane under the influence of a force such that its linear momentum is $\mathrm{P}^{-}(\mathrm{t})=\mathrm{A}[\mathrm{i} \cos (\mathrm{kt})-\mathrm{j} \sin (\mathrm{kt})]\( where \)\mathrm{A}$ and \(\mathrm{k}\) are constants. The angle between the force and momentum is (A) \(0^{\circ}\) (B) \(30^{\circ}\) (C) \(45^{\circ}\) (D) \(90^{\circ}\)
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