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  • A projectile of mass M is fired so that the horizontal range is 4 km At the highest point

IMG_20180809_174539_554.JPG A projectile of mass M is fired so that the horizontal range is 4 km. At the highest point the projectile explodes in two parts of masses M/4 and 3M / 4 respectively and the heavier part starts falling down with zero initial speed. The horizontal range (distance from point of fringe) of the lighter part is (A) 16 km (B) 8 km (C) 10km (D) 12 km

Answers (1)

@Roshni Afrin 

The heavier part has no horizontal momentum after collision as it falls vertically down.   Let the lighter part have a horizontal velocity v.

     M* u cos? = 3M/4 * 0 + M/4 * v
     v = 4 u cos?

The time duration t before the lighter mass reaches ground:
       H = 0 t + 1/2 g t^2
       t^2 = 2 H/g = (2,000/g) Tan?                                        (since H=Rtan? /4)
       t = 20 √(5 tan? /g)  sec

Horizontal Distance traveled by lighter mass after explosion = v t
       vt  = 80 u cos? √5tan?/g)
            =  80 u √(5sin? cos?/g) 
            = 40 √(10 u² sin2?/g)

= 40  √[10*4000/g]                                      (since u² sin2?/g=4000m ;given)

=8Km

so total range of lighter part =10Km

 

Posted by

Safeer PP

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