Question

A particle moves in the xy-plane under the action of force F such that the components of its linear momentum p at any time t are \[{p_x} = 2\cos t\], \[{p_y} = 2\sin t\]. The angle between F and p at time t is
A. \[90^\circ \]
B. \[0^\circ \]
C. \[180^\circ \]
D. \[30^\circ \]

Answer 1

Hint: Calculate the magnitude of the linear momentum. Then express the kinetic energy of a particle in terms of linear momentum. Find the work done of the particle which is equal to the change in the kinetic energy of the particle. Power is equal to work done per unit time and it is also the dot product of force and velocity. Multiply and divide the equation by the mass of the particle and express the angle between force and momentum.

Formula used:
Kinetic energy, \[K = \dfrac{{{p^2}}}{{2m}}\]
Here, p is the linear momentum and m is the mass of the particle.
Power, \[P = \dfrac{W}{t}\]
Here, W is the work done and t is the time.

Complete step by step answer:
We have given the horizontal and vertical component of the linear momentum p. We can calculate the magnitude of the linear momentum as follows,
\[p = \sqrt {p_x^2 + p_y^2} \]
Here, \[{p_x}\] is the horizontal component of the linear momentum and \[{p_y}\] is the vertical component of the linear momentum.
Substituting \[{p_x} = 2\cos t\] and \[{p_y} = 2\sin t\] in the above equation, we get,
\[p = \sqrt {{{\left( {2\cos t} \right)}^2} + \left( {{{\left( {2\sin t} \right)}^2}} \right)} \]
\[ \Rightarrow p = \sqrt {4\left( {{{\sin }^2}t + {{\cos }^2}t} \right)} \]

We know that, \[{\sin ^2}t + {\cos ^2}t = 1\]. Therefore, the above equation becomes,
\[p = \sqrt 4 \]
\[ \Rightarrow p = 2\]

We can express the kinetic energy of the particle as follows,
\[K = \dfrac{{{p^2}}}{{2m}}\]
Here, m is the mass of the particle.

Substituting 2 for p in the above equation, we get,
\[K = \dfrac{{{{\left( 2 \right)}^2}}}{{2m}}\]
\[ \Rightarrow K = \dfrac{2}{m}\]
Therefore, we can see that the kinetic energy of the particle is constant.

We know that the work done is the change in kinetic energy of the particle. Therefore, the work done is surely equal to zero. We also know that the power is expressed as,
\[P = \dfrac{W}{t}\]
Here, W is the work done and t is the time.

Since the work done is zero, the power also must be zero. Therefore, \[P = 0\].
We have the expression for power is,
\[P = \vec F \cdot \vec v = 0\]
Multiplying and dividing the above equation by the mass of the particle, we get,
\[\dfrac{{\vec F \cdot \left( {m\vec v} \right)}}{m} = 0\]
\[ \Rightarrow \vec F \cdot \left( {m\vec v} \right) = 0\]

But we know that the linear momentum of the particle is equal to, \[\vec p = m\vec v\]. Therefore, the above equation becomes,
\[\vec F \cdot \vec p = 0\]
\[ \Rightarrow FP\cos \theta = 0\]
\[ \Rightarrow \cos \theta = 0\]
\[ \Rightarrow \theta = {\cos ^{ - 1}}0\]
\[ \therefore \theta = 90^\circ \]

So, the correct answer is option A.

Note:One can also solve this question by taking the dot product of the vectors F and p. The vector dot product of these two vectors can be expressed as, \[\vec F\vec p = Fp\cos \theta \]. The force vector is the sum of its horizontal component and vertical component. You will get \[\cos \theta \] equal to 0.