Physics-
General
Easy

Question

From an inclined plane two particles are projected with same speed at same angle theta, one up and other down the plane as shown in figure. Which of the following statements left parenthesis s right parenthesis is/are correct?

  1. The time of flight of each particle is the same.    
  2. The particles will collide the plane with same speed    
  3. Both the particles strike the plane perpendicularly    
  4. The particles will collide in mid air if projected simultaneously and time of flight of each particle is less than the time of collision    

The correct answer is: The time of flight of each particle is the same.


    Here, alpha equals 2 theta comma beta equals theta

    Time of flight of A is, T subscript 1 end subscript equals fraction numerator 2 u sin invisible function application left parenthesis alpha minus beta right parenthesis over denominator g blank c o s blank beta end fraction
    equals fraction numerator 2 u sin invisible function application left parenthesis 2 theta minus theta right parenthesis over denominator g blank c o s blank theta end fraction equals fraction numerator 2 u over denominator g end fraction t a n invisible function application theta
    Time of flight of B is, T subscript 2 end subscript equals fraction numerator 2 u sin invisible function application theta over denominator g cos invisible function application theta end fraction equals fraction numerator 2 u over denominator g end fraction tan invisible function application theta
    So, T subscript 1 end subscript equals T subscript 2 end subscript. The acceleration of both the particles is g downwards. Therefore, relative acceleration between the two is zero or relative motion between the two is uniform. The relative velocity of A w.r.t. B is towards A B blank, therefore collision will take place between the two in mid air.

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