Respuesta :
The ratio of the initial speed of the 9-mm bullet to the speed of the 0.44-caliber bullet is; v₀₁/v₀₂ = 0.62
What is the ratio of the initial speed?
We are given;
Mass of pendulum; M = 10
Mass of bullet; m₁ = 6
Mass of caliber bullet; m₂ = 12
Maximum angular displacement due to bullet; θ₁ = 4.3
Maximum angular displacement due to caliber bullet; θ₂ = 10.1
The speed of the bullet is calculated as;
[tex]v_{01} = (1 + \frac{M}{m_{1} })\sqrt{2gL(1 - cos \theta_{1} ) }[/tex]
where;
L is is the length of the pendulum
Substituting the relevant values gives;
[tex]v_{01} = (1 + \frac{10}{6})\sqrt{2gL(1 - cos 4.3 ) }[/tex] -------(eq 1)
Likewise, speed of the caliber is;
[tex]v_{02} = (1 + \frac{M}{m_{2} })\sqrt{2gL(1 - cos \theta_{2} ) }[/tex]
Plugging in the relevant values gives;
[tex]v_{02} = (1 + \frac{10}{12})\sqrt{2gL(1 - cos {10.1} ) }[/tex] -------(eq 2)
When we divide eq 1 by eq 2 and simplify, we have;
v₀₁/v₀₂ = (2.67 * 0.053)/(1.84 * 0.124)
Ratio of speed is; v₀₁/v₀₂ = 0.62
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