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In a ballistic pendulum an object of mass m is fired with an initial speed v_0 at a pendulum bob. The bob has a mass M, which is suspended by a rod of length L and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement (theta) as shown

An experiment is done to compare the initial speed of bullets fired from different handguns: a 9 mm and a .44 caliber. The guns are fired into a 10-kg pendulum bob of length L. Assume that the 9-mm bullet has a mass of 6 g and the .44-caliber bullet has a mass of 12 g . If the 9-mm bullet causes the pendulum to swing to a maximum angular displacement of 4.3degrees and the .44-caliber bullet causes a displacement of 10.1degrees , find the ratio of the initial speed of the 9-mm bullet to the speed of the .44-caliber bullet, (v_0)_9/(v_0)_44.

My Solution

I already found the inital speed of the fired object: v = ((m+M)/m)*sqrt(2gL[1 - cos(theta)]). So I am given m (bullet mass), M (block's mass), g (gravity), and theta. I plugged those into the equation and got (v_0)_44=((10kg + .012)/10kg)*sqrt(2*9.8*L[1 - cos 10.1]) and (v_0)_9=((10kg + .006)/10kg)*sqrt(2*9.8*L[1 - cos 4.3]).

Am I on the right track? What happens after?

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# Homework Help: Ballistic Pendulum displacement

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