Question

A particle has two velocity v1 andv2 .its resultant velocity is equal to v1 in magnitude. Find the angle which the new resultant makes with v2 when v1 is doubled.

Answer 1

Let the magnitudes of velocities $v_{1}, v_{2}$ be $p, q$ respectively.
And the angle between them be $\theta$.
From the data provided,
[tex]p^{2}=p^{2}+q^{2}+2pq\cos\theta \\ q+2p\cos \theta=0[/tex].
Now, the angle made by new resultant with $v_{2}$ when $v_{1}$ is doubled is$\tan\phi=\frac{2p\sin\theta}{q+2p\cos\theta}$,
Since denominator is 0, angle made by resultant is 90°.

Answer 2

See the diagram for the way vectors are added and the angle that resultant makes.

Resultant vector $\vec{V}\ of\ \vec{V_1}\ and\ \vec{V_1}$ with an angle Ф between them is given by the law of vector addition :

   V² = V₁² + V₂²  + 2 V₁ V₂ Cos Ф        --- (1)
       = V₁²      given

=>   V₂ =  - 2 V₁ Cos Ф              ---- (2)

    Angle δ between the resultant vector V  and V₂ is given by :

$Tan\delta = \frac{V_1\ Sin\ \phi }{V_2+V1\ Cos\phi}=\frac{-Sin\phi}{Cos\phi}=-tan(-\phi)\\\\Hence,\ \delta=-\phi,\ \ \ or,\ \ \pi-\phi.$        ----- (3)

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Now , the magnitude of $\vec{V_1}$ is doubled.  V₂ remains same.

  resultant V² = (2V₁)² + V₂² + 2 * 2V₁* V₂ * Cos Ф
                   = 4 V₁² + (-2V₁ Cos Ф)² + 4 V₁ (-2V₁ CosФ) Cos Ф
                   =  4 V₁² ( 1 - Cos² Ф)  = 4 V₁² Sin² Ф
          | V |  =  | 4 V₁ Sin Ф |    magnitude of the resultant      -----  (4)

   Angle δ'  that Resultant vector $\vec{V}$ makes with V₂ is:
 
$Tan\delta'=\frac{(2V_1)Sin\phi}{V_2+(2V_1)Cos\phi}=\frac{2V_1Sin\phi}{-2V_1Cos\phi+2V_1Cos\phi}$      ----- (5)
    
We see that the denominator is 0.  It means that tan δ' is infinity.  Hence  the angle δ' that the resultant makes with the velocity V₂  is  π/2.