Question

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) \(90^{\circ}\)

Short answer

The angle between the force and momentum at the specific time when \(kt=\frac{\pi}{4}\) could not be found among given options (A), (B), (C), or (D). In general, it would vary, making a straight answer impossible.

Step-by-step solution

Determine the force using momentum expression

Given the linear momentum of a particle P(t) is: \(P(t) = A[i \cos(kt) - j \sin(kt)]\), where A and k are constants. We know that force (F) is the rate of change of linear momentum: \(F = \frac{dP}{dt}\). So let's differentiate P(t) with respect to time (t) to find F(t). 2. Differentiate P(t) with respect to time

Compute the force F(t) by differentiating P(t)

Differentiating P(t) with respect to time (t), we get: \(F(t) = \frac{d}{dt}[A(i \cos(kt) - j \sin(kt))]\). Applying the product rule separately to each term: \[F(t) = A\left[- i k \sin(kt) - j k \cos(kt)\right]\] 3. Find the dot product of the force and momentum vectors

Calculate the dot product of F(t) and P(t)

To determine the angle between F(t) and P(t), we need to calculate the dot product. The dot product is given by: \[F(t) \cdot P(t) = |F(t)| \times |P(t)| \times \cos(\theta)\] We have F(t) = \[A\left[- i k \sin(kt) - j k \cos(kt)\right]\] and P(t) = \(A[i \cos(kt) - j \sin(kt)]\). Calculating the dot product, we get: \[ (F\cdot P) = A\left[- ik \sin(kt) - jk \cos(kt)\right] \cdot A[i \cos(kt) - j \sin(kt)] \\ = A^2k \left[- \sin(kt)\cos(kt) - \sin(kt)\cos(kt)\right] = -2A^2k \sin(kt) \cos(kt) \] 4. Find magnitudes of F(t) and P(t)

Compute magnitudes of force and momentum vectors

Next step is to find the magnitudes of F(t) and P(t). The magnitudes are calculated as follows: Magnitude of P(t): \[|P(t)| = |A[i \cos(kt) - j \sin(kt)]|= A\sqrt{(\cos^2(kt) + \sin^2(kt))}= A\] Magnitude of F(t): \[|F(t)| = |A[-ik\sin(kt) - jk\cos(kt)]|= Ak\sqrt{(\sin^2(kt) + \cos^2(kt))}= Ak\] 5. Find the angle between the force and momentum

Calculate the angle between F(t) and P(t) using dot product

We have computed the dot product \[F(t) \cdot P(t) = -2A^2k \sin(kt) \cos(kt)\] and magnitudes |F(t)| = Ak, |P(t)| = A. Now, we can calculate the angle θ using the dot product formula: \[ \cos(\theta) = \frac{F(t) \cdot P(t)}{|F(t)| \times |P(t)|} = \frac{-2A^2k\sin(kt)\cos(kt)}{A\cdot A\cdot k} = -2\sin(kt)\cos(kt) \] Notice that the expression for cosine of the angle doesn't depend on the constants A and k. Now, we need to find the angle θ, for which cos(θ) is equal to the above expression. We observe that the case when sin(kt)=cos(kt), i.e., two angles are equal, corresponds to \(kt=\frac{\pi}{4}\), i.e., sin(kt)=cos(kt)=\(\frac{1}{\sqrt{2}}\). In this case: \[ \cos(\theta) = -2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)=-\frac{1}{2} \] This means that \(\theta = \arccos(-\frac{1}{2})\), which corresponds to an angle of 120°. None of the given options lists 120°, which could be justified by asking not for the general angle, but for specific points in time where the dot product could momentarily differ from the general case. To rigorously justify the answer, a more thorough analysis would be needed. Thus, the angle between the force and momentum at the specific time when \(kt=\frac{\pi}{4}\) could not be found among given options (A), (B), (C), or (D). In general, it would vary, making a straight answer impossible.