Question

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

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

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: \(\frac{dP}{dt} = \frac{d}{dt}(a + bt^2)\)

2) Differentiate the Constants

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

3) Differentiate the Second Term

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

4) Combine the Results

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

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: \(F = 2bt\)

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