Veocity of the bullet before collision:

${\mathit{v}}_{\mathbf{0}}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{M}}{\mathbf{m}}\mathbf{)}\sqrt{\mathbf{2}\mathbf{gL}\mathbf{(}\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathbf{cos\theta}\mathbf{)}}$

M = 10kg

m_{9.0} = 0.006 kg

m_{.44} = 0.012 kg

θ_{9.0} = 4.3°

θ_{.44} = 10.1°

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 θ as shown (Figure 1)

The expression for the initial velocity v_{0} of the bullet in terms of m, M, L, θ, and the acceleration due to gravity, *g is *

An experiment is done to compare the initial speed of bullets fired from different handguns: a 9.0 mm and a .44 caliber. The guns are fired into a 10-kg pendulum bob of length *L*. Assume that the 9.0-mm bullet has a mass of 6.0 g and the .44-caliber bullet has a mass of 12 g . If the 9.0-mm bullet causes the pendulum to swing to a maximum angular displacement of 4.3?and the .44-caliber bullet causes a displacement of 10.1? , find the ratio of the initial speed of the 9.0-mm bullet to the speed of the .44-caliber bullet, (*v*_{0})_{9.0}/(*v*_{0})_{44}.

Express your answer numerically.

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