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Question: A particle moves in the \(x - y\) plane under the influence of a force such that its linear momentum...

A particle moves in the x−yx - y plane under the influence of a force such that its linear momentum is p→(t)=A(i^cos⁡(kt)−j^sin⁡(kt))\overrightarrow p \left( t \right) = A\left( {\hat i\cos \left( {kt} \right) - \hat j\sin \left( {kt} \right)} \right) where AA and kk are constants. The angle between the force and momentum is
(A) 0∘0^\circ
(B) 30∘30^\circ
(C) 45∘45^\circ
(D) 90∘90^\circ

Explanation

Solution

Momentum of a body is nothing but the product of mass and velocity of a body. It is a vector quantity that has both magnitude and direction variables.
p→=mv\overrightarrow p = mv
However, newton’s second law of motion states that the rate of change of momentum of a body over time is directly proportional to the force applied and also occurs in the same direction as the applied force.
F=dpdtF = \dfrac{{dp}}{{dt}} .
Formula used: We will be using the formula F=dpdtF = \dfrac{{dp}}{{dt}} which is the mathematical expression for Newton’s second law of motion, where FF is the force applied on the body, pp is the momentum of the body, and tt is the time it takes to change, thus making dpdt\dfrac{{dp}}{{dt}} the rate of change of momentum.

Complete Step by Step answer:
We know that momentum is a vector quantity that has both magnitude and direction. We also know that mathematically momentum is nothing but the product of mass and velocity.
p→=mv\overrightarrow p = mv
In this the mass is a scalar quantity while the velocity is the vector quantity thus the direction of momentum depends on the direction of velocity. We also know from Newton’s Second law of motion that force is directly proportional to the rate of change of momentum of a body.
F=dpdtF = \dfrac{{dp}}{{dt}}
From the given problem we have the momentum of the particle as a function of time,
p→(t)=A(i^cos⁡(kt)−j^sin⁡(kt))\overrightarrow p \left( t \right) = A\left( {\hat i\cos \left( {kt} \right) - \hat j\sin \left( {kt} \right)} \right) where A,kA,k are constants.
We are required to find the angle between the momentum and force vector.
Using Newton's Second law of motion, we need to differentiate the momentum with respect to time to get an expression of force.
Differentiating the function p→(t)\overrightarrow p \left( t \right) with respect to tt
F→=dpdt=ddt[A(i^cos⁡(kt)−j^sin⁡(kt))]\overrightarrow F = \dfrac{{dp}}{{dt}} = \dfrac{d}{{dt}}\left[ {A\left( {\hat i\cos \left( {kt} \right) - \hat j\sin \left( {kt} \right)} \right)} \right]
=Ak[(−i^sin⁡(kt)−j^cos⁡(kt))]= Ak\left[ {\left( { - \hat i\sin \left( {kt} \right) - \hat j\cos \left( {kt} \right)} \right)} \right]
[Since d(sin⁡x)dx=cos⁡x\dfrac{{d\left( {\sin x} \right)}}{{dx}} = \cos x and d(cos⁡x)dx=−sin⁡x\dfrac{{d\left( {\cos x} \right)}}{{dx}} = - \sin x]
Now that we have a function of force with respect to time, to find the angle between force and momentum vector, we need to find the dot product of the two vectors, F→.p→\overrightarrow F .\overrightarrow p
F→.p→=Ak[(−i^sin⁡(kt)−j^cos⁡(kt))].[A(i^cos⁡(kt)−j^sin⁡(kt))]\overrightarrow F .\overrightarrow p = Ak\left[ {\left( { - \hat i\sin \left( {kt} \right) - \hat j\cos \left( {kt} \right)} \right)} \right].\left[ {A\left( {\hat i\cos \left( {kt} \right) - \hat j\sin \left( {kt} \right)} \right)} \right]
Solving the dot product, we get,
F→.p→=Ak[(−sin⁡(kt)cos⁡(kt)+cos⁡(kt).sin⁡(kt))]\overrightarrow F .\overrightarrow p = Ak\left[ {\left( { - \sin \left( {kt} \right)\cos \left( {kt} \right) + \cos \left( {kt} \right).\sin \left( {kt} \right)} \right)} \right]
F→.p→=Ak(0)\overrightarrow F .\overrightarrow p = Ak\left( 0 \right) [since −sin⁡(kt)cos⁡(kt)+cos⁡(kt).sin⁡(kt)=0 - \sin \left( {kt} \right)\cos \left( {kt} \right) + \cos \left( {kt} \right).\sin \left( {kt} \right) = 0]
⇒F→.p→=0\Rightarrow \overrightarrow F .\overrightarrow p = 0
We also know that the dot product is the product of the respective resultant vectors and the cosine of the angle between them.
F→.p→=Fpcos⁡θ\overrightarrow F .\overrightarrow p = Fp\cos \theta
Since F→.p→=0\overrightarrow F .\overrightarrow p = 0, we know Fpcos⁡θ=0Fp\cos \theta = 0, so cos⁡θ=0\cos \theta = 0 .
The cosine values are null when, θ=90∘\theta = 90^\circ .

Thus, the angle between momentum and force is θ=90∘\theta = 90^\circ . Hence the correct answer is option D

Note: We have calculated the dot product Ak[(−i^sin⁡(kt)−j^cos⁡(kt))].[A(i^cos⁡(kt)−j^sin⁡(kt))]=Ak[(−sin⁡(kt)cos⁡(kt)+cos⁡(kt).sin⁡(kt))]Ak\left[ {\left( { - \hat i\sin \left( {kt} \right) - \hat j\cos \left( {kt} \right)} \right)} \right].\left[ {A\left( {\hat i\cos \left( {kt} \right) - \hat j\sin \left( {kt} \right)} \right)} \right] = Ak\left[ {\left( { - \sin \left( {kt} \right)\cos \left( {kt} \right) + \cos \left( {kt} \right).\sin \left( {kt} \right)} \right)} \right]This is because the dot products of
i^.i^=1=j^.j^\hat i.\hat i = 1 = \hat j.\hat j and i^.j^=0=j^.i^\hat i.\hat j = 0 = \hat j.\hat i.