Question

A particle of mass m is made to move with uniform speed u along the perimeter of a regular polygon of n sides. What is the magnitude of impulse applied by the particle at each corner of the polygon?

Answer 1

Hint: Apply the subtraction for two vectors by the law of parallelogram i.e., \[R=\sqrt{{{P}^{2}}+{{Q}^{2}}-2PQ\cos \theta }\] and take the angle between two sides of polygon \[\dfrac{2\pi }{n}\]

Complete step by step answer:
Given that the particle is moving with uniform speed u along the regular polygon of n sides. Drawing a rough figure will lead to following diagram, (assuming polygon with 7 sides)

Therefore,
The initial momentum is given by, \[{{p}_{i}}=mu\left( \uparrow \right)\] and
The final momentum is given by, \[{{p}_{f}}=mu\left( -\uparrow \right)\]

Now, impulse is given by,

\[\overrightarrow{J}=\overrightarrow{\Delta p}=\overrightarrow{{{p}_{f}}}-\overrightarrow{{{p}_{i}}}\]
\[\overrightarrow{J}=mu\left( -\uparrow \right)-mu\left( \uparrow \right)\]
As angle between the \[\overrightarrow{{{p}_{i}}}\] and \[\overrightarrow{{{p}_{f}}}\] is \[\theta =\dfrac{2\pi }{n}\]

Also given that, \[{{p}_{i}}={{p}_{f}}=p=mu\]

Now, by the law of parallelogram,

\[J=\sqrt{{{p}^{2}}+{{p}^{2}}-2{{p}^{2}}\cos \left( \dfrac{2\pi }{n} \right)}\]
\[J=\sqrt{2{{p}^{2}}-2{{p}^{2}}\cos \left( \dfrac{2\pi }{n} \right)}\]
\[J=\sqrt{2{{p}^{2}}\left( 1-\cos \left( \dfrac{2\pi }{n} \right) \right)}\]

Using half-angle formula of trigonometry,

\[J=\sqrt{4{{p}^{2}}\left( {{\sin }^{2}}\left( \dfrac{\pi }{n} \right) \right)}\]
\[J=2p\sin \left( \dfrac{\pi }{n} \right)\]

Substituting \[p=mu\], we get

\[J=2mu\sin \left( \dfrac{\pi }{n} \right)\]
Hence, the magnitude of impulse\[\left( J \right)\] applied by the particle at each corner of the polygon is \[2mu\sin \left( \dfrac{\pi }{n} \right)\].

Additional Information:
Momentum: The momentum of an object is the product of its mass and velocity. An object can have a large momentum by having a large mass or a large velocity. Momentum is a vector quantity and it is important to pay attention to the signs of the components of momentum.

Impulse: A force of short duration is an impulsive force. The impulse (J) that this force delivers to an object is the area under the force vs. time graph. For time dependent forces, impulse and momentum are often more useful than Newton’s laws.

Note: Students should keep in mind that the momentum is a vector quantity, so all the operations such as addition and subtraction are governed under vector addition and multiplication. Along with this, students need to understand the law of parallelogram for vector addition and subtraction so that they can do this type of questions easily.
Students often get confused with the impulse of the polygon, please remember impulse in case of a polygon is along the line joining the vertex and centre of the polygon.