a bullet of mass m1 is fired horizontally with a speed v into the bob of a balli

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Explanation / Answer

See when the bullet hit the bob it gets stuck to it and since the horizontal force on the system is zero the momentum in the Horizontal direction is conserved. By momentum conservation we have m1v = (m1+m2)v1 =>v = (m1+m2)/m1 Now this is the speed of the mass + bullet system in the bottom of the circle. For the system to make a complete circle it must have a velocity at the top such that the sting doesn't slack. Now Let the length of string of be l So at the top most point let the velocity of the bob be V (m1+m2)V^2/l = (m1+m2)g => v= root(gl) Now by energy conservation between the top and bottom point we have .5(m1+m2)v1^2 = 2(m1+m2)gl + .5(m1+m2)V^2 =>.5v1^2 = 2gl + .5gl => v1^2 = 5gl v1 = root(5gl) The velocity of bullet required = (m1+m2)root(5gl)/m1

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