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  • Brainly User
2015-04-23T16:54:03+05:30
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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2015-04-23T21:21:00+05:30

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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)

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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.

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