Question

A particle is moving in the x-y plane under the influence of a force such that its linear momentum is
P(t)=A(cos(kt)i-sin(kt)j), A and k are constants. The angle b/w force and linear mom is ????

Answer 1

Answer on Question #40825, Physics, Mechanics Question: A particle is moving in the x-y plane under the influence of a force such that its linear momentum is $\vec{P}(t) = A(\cos(kt)\vec{i} - \sin(kt)\vec{j})$, A and k are constants. The angle b/w force and linear mom is ???? Solution. First we find the coordinates of vector of the force.

$$\vec {F} = \frac {d \vec {p}}{d t} = - A k (\sin (k t) \vec {i} + \cos (k t) \vec {j})$$

Now the angle is

$$(\vec {p} \cdot \vec {F}) = | \vec {p} | | \vec {F} | \cos \alpha$$ $$\alpha = \arccos \left(\frac {(\vec {p} \cdot \vec {F})}{| \vec {p} | | \vec {F} |}\right) = \arccos \left(\frac {- 2 A ^ {2} k (\cos (k t) \sin (k t))}{- 2 A ^ {2} k (1 - \sin^ {2} (k t) \cos^ {2} (k t))}\right) = \arccos \left(\frac {\sin (2 k t) / 2}{1 - \sin^ {2} (2 k t) / 4}\right)$$