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A ball of mass 0.2 kg …

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A ball of mass 0.2 kg rest on a vertical pole of height 5 m. A bullet of mass 0.01 kg travelling with a velocity v m/sec in a horizontal direction,hits the center of ball. After the collision, the ball and the bullet travel independently. The ball hits the ground at a distance of 20m and the bullet at a distance of 100m from the foot of the post. Find the initial velocity of the bullet...

    • Posted by Cluster Muster 3 years, 10 months ago

    CBSE > Class 11 > Physics
    • 1 answers

    Nikku Ad 3 years, 10 months ago

    Let the mass of the bullet be m and mass of ball be M Initially the ball is at a height 5m and t rest,the only acceleration is due to gravity therefore applying equation of motion S=ut+½gt² 5=0+½(10)t² therefore t=1sec So V(ball)=20m/s V(bullet)=100m/s So by collision M*V(ball:final)+m*V(bullet:final)= M*V(ball:initial)+m*V(bullet: initial) 0.01*V(bulletfinal)= 0.01*100+0.20*20 V=500m/s

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