In a ballistic pendulum an object of mass m is fired with an initial speed v0 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 ? as shown (Figure 1)

7 0 0 Mm 1-

Find an expression for v0, the initial speed of the fired object.

Express your answer in terms of some or all of the variables m, M, L, and ? and the acceleration due to gravity, g.

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, (v0)9.0/(v0)44.

For the expression for the initial speed of the projectile, we can separate the problem in two phases. The first one is the moment before and after the impact. The second phase is the rising of the ballistic pendulum.

First Phase: Impact

In the process of the impact, the net external forces acting in the system bullet-pendulum are null. Therefore the linear momentum remains even (Conservation of linear momentum). This means:

[tex]P_0=P_f\\v_0m=v_i(m+M)\\v_0=v_i\frac{m+M}{m}[/tex] (1)

Second Phase: pendular movement

After the impact, there isn't any non-conservative force doing work in al the process. Therefore the mechanical energy remains constant (Conservation Of Mechanical Energy). Therefore:

[tex]Em_i=Em_f\\\frac{1}{2}mv^2_i=mgH\\v_i=[2gH]^\frac{1}{2}[/tex] (2)

The height of the pendulum respect L and θ is:

[tex]H=L(1-cos(\theta))[/tex] (3)

Using equations (1),(2) and (3):

[tex]\displaystyle v_0=\frac{M+m}{m}(2gL[1-cos(\theta)]^{\frac{1}{2}})[/tex] (4)

The initial speed ratio for the 9.0mm/44-caliber bullet is obtained using equation (4):

[tex]\displaystyle \frac{v_{9mm}}{v_{44}} =\frac{(M+m_{9mm})m_{44}}{(M+m_{44})m_{9}}(\frac{1-cos(\theta_{9mm})}{1-cos(\theta_{44})} )^{\frac{1}{2}}=1.773[/tex]

jacqeroyer jacqeroyer · Answer 2 · 2021-11-19T00:07:02+01:00

Answer:

[tex]\frac{(M+m)}{m}\sqrt{2gL(1-cos\theta)}[/tex]

Explanation:

The first half can be determined by Conservation of Momentum:

[tex]mv_{o} =(M+m)v[/tex]

solve for the initial speed, then use Conservation of Energy to solve for the height that the pendulum swings to. The relationship here between the equation that we just solved for is that the final speed of the two objects after they stick together is the kinetic energy for the "initial" in the Conservation of Energy equation:

[tex]\frac{1}{2}(M+m)v_o=(M+m)gh[/tex]

The mass cancels on both sides, so the only thing left to determine is the height of the pendulum in terms of L and [tex]\theta[/tex], which can be accomplished via trig:

Think of the initial and final position of the pendulum as two sides of the right triangle, with an angle θ in between. The triangle can be turned into a right triangle be connecting the final displacement of the pendulum to the initial location of the string. This shows us that the vertical displacement of the pendulum is h and that L-h is the length of the vertical leg. Therefore:

[tex]cos\theta=\frac{L-h}{L}[/tex] , so h=L-Lcos(theta)

We can substitute this into [tex]\frac{1}{2}(M+m)v_o=(M+m)gh[/tex] to see that [tex]\frac{1}{2}v_o=gL(1-cos\theta)[/tex]

Then solve for the velocity and substitute this into the Conservation of Momentum formula and simplify:

[tex]\frac{(M+m)}{m}\sqrt{2gL(1-cos\theta)}[/tex]

To solve the second part, convert to SI units and plug the values into the above equation