Question

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

Answer 1

Hint: The force is defined as the push or pull of an object that results in the change in velocity of that object. Linear momentum is defined as the product of the mass of an object and the velocity of that object. Therefore, we will calculate the angle between the force and linear momentum by using the dot product of force and linear momentum.

Complete step by step answer:
As given in the question, the linear momentum of a particle is $\vec p(t) = \,A\left( {i\,\cos (kt) - j\,\sin (kt)} \right)$
Now, differentiating both sides of the above equation, we get
$\dfrac{{dp}}{{dt}} = A\left( { - i\sin (kt) \times k - j\cos (kt) \times k} \right)$
$ \Rightarrow \,\dfrac{{dp}}{{dt}} = A\left( { - ik\,\sin (kt) - jk\,\cos (kt)} \right)$

Now, the force of an object according to newton’s law is defined as the ratio of the change in momentum of a particle to the change in time and is given by,
$F = \dfrac{{dp}}{{dt}}$
Therefore, putting the value of $\dfrac{{dp}}{{dt}}$ , we get
$F = A\left( { - ik\,\sin (kt) - jk\,\cos (kt)} \right)$
Now, let the angle between the force and the linear momentum is $\theta $ .
Therefore, the angle between the force and linear momentum is
$\cos \,\theta = \dfrac{{F.\vec p}}{{\left| F \right|\left| {\vec p} \right|}}$

Now, putting the values of $F$ and $\vec p$ , we get
$\cos \theta = \dfrac{{A\left( { - ik\,\sin (kt) - jk\,\cos (kt)} \right).A\left( {i\,\cos (kt) - j\,\sin (kt)} \right)}}{{\sqrt {{A^2}{{\left( { - ik\,\sin (kt) - ik\,\cos (kt)} \right)}^2} - {{\left( {{A^2}\left( {i\,\cos (kt) - j\,\sin (kt)} \right)} \right)}^2}} }}$
$ \Rightarrow \,\cos \theta = \dfrac{{{A^2}\left[ {\left( { - i.ik\,\sin (kt)\,\cos (kt)} \right) + \left( {j.j\sin (kt)\,\cos (kt)} \right)} \right]}}{{A\sqrt {\left( {{i^2}{k^2}\,{{\sin }^2}(kt) + {j^2}{k^2}{{\cos }^2}(kt) + i.j{k^2}\sin (kt)\,\cos (kt)} \right) - \left( {{i^2}\,{{\cos }^2}(kt) + {j^2}\sin (kt) + ij\,\cos (kt)\,\sin (kt)} \right)} }}$
$ \Rightarrow \,\cos \theta = \dfrac{{{A^2}\left[ {\left( { - k\,\sin (kt)\,\cos (kt)} \right) + \left( {k\sin (kt)\,\cos (kt)} \right)} \right]}}{{A\sqrt {\left( {{i^2}{k^2}\,{{\sin }^2}(kt) + {j^2}{k^2}{{\cos }^2}(kt) + i.j{k^2}\sin (kt)\,\cos (kt)} \right) - \left( {{i^2}\,{{\cos }^2}(kt) + {j^2}\sin (kt) + ij\,\cos (kt)\,\sin (kt)} \right)} }}$
$ \Rightarrow \,\cos \theta = 0$
$ \therefore \,\theta = 90^\circ $
Therefore, the angle between the force and the linear momentum is $90^\circ $ .

Hence, (D) is the correct option.

Note:Newton’s second law of motion is also applicable to the conservation of linear momentum. When the net force acting on the body will be zero, the momentum of the body is constant. Also, the net force acting on the body is equal to the rate of change of the momentum. Now, when the mass of an object increases or decreases there will be a change in the momentum of an object. Therefore, the momentum of an object changes with the change in the mass of an object.