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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



Let the magnitudes of velocities v_{1}, v_{2} be p, q respectively.
And the angle between them be \theta.
From the data provided,
p^{2}=p^{2}+q^{2}+2pq\cos\theta \\
q+2p\cos \theta=0.
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°.
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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)

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.

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