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Question

A ball of mass 0.2 blank k g rests on a vertical post of height 5 blank m. A bullet of mass 0.01 blank k g, travelling with a velocity V blank m divided by s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 blank mand the bullet at a distance of 100 blank mfrom the foot of the post. The initial velocity V of the bullet is

  1. 250 blank m divided by s    
  2. 250 square root of 2 blank m divided by s    
  3. 400 blank m divided by s    
  4. 500 blank m divided by s    

The correct answer is: 500 blank m divided by s


    R equals u square root of fraction numerator 2 h over denominator g end fraction end root rightwards double arrow 20 equals V subscript 1 end subscript square root of fraction numerator 2 cross times 5 over denominator 10 end fraction end root and 100 equals V subscript 2 end subscript square root of fraction numerator 2 cross times 5 over denominator 10 end fraction end root
    rightwards double arrow V subscript 1 end subscript equals 20 blank m divided by s comma blank V subscript 2 end subscript equals 100 blank m divided by s
    Applying momentum conservation just before and just after the collision open parentheses 0.01 close parentheses open parentheses V close parentheses equals open parentheses 0.2 close parentheses open parentheses 20 close parentheses plus left parenthesis 0.01 right parenthesis left parenthesis 100 right parenthesis
    V equals 500 blank m divided by s

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