What is the Initial Speed Formula for a Ballistic Pendulum?

[dandirom]

[SOLVED] Initial Speed formula

Homework Statement

A bullet of mass m = 0.01 kg is fired into a ballistic pendulum of mass M = 2.0 kg as shown in Figure 1. The bullet remains in the block after the collision and the whole system rises to a maximum height of h = 8 cm. What is the bullet’s initial speed?

Homework Equations

I'm really stumped with what equation to apply to this problem. I want to work it out on my own but I just need help with the what equation to apply.

The Attempt at a Solution

I've tried the formulas for initial speed.

 

[mbrmbrg]

You need to split up this scenario into two sections. In the first part, momentum is conserved. In the second, energy is conserved. Is that about the size hint you were looking for?

Also, what "equation for initial speed" did you utilize?

 

[dandirom]

oh ok. thank you.

it's a multiple choice question and most formulas I found already had an initial speed given so I tried substitution.

I tried to calculate the speed in m/s if both masses travel 8cm.

then i multiplied the masses w/ the velocity.

0.01 X velocity =
2.0 X velocity =
Total
then i got the sum of both answers so
0.01 x v = total

V = (Total / 0.01)
I don't know if I used the right formula though.

 

[cristo]

Your notation is rather ambiguous. Try to get used to using symbols once and only once; this will prevent confusion.

You should use conservation of energy here. Consider the energy before as the sum of the kinetic energies of the two objects, and the total energy after will be the potential energy of the bullet/block.

 

[HallsofIvy]

Looks to me like "total energy" is the way to go. Taking the base point for potential energy at the initial height of the pendulum, bullet and pendulum have 0 potential energy and only the bullet has kinetic energy. The total energy of the system is the kinetic energy of the bullet: [itex]\frac{1}{2}mvp². You know m= 0.1 kg but v is unknown. When the bullet, pendulum combination reaches its maximum height, neither has kinetic energy but you can calculate their potential energy, mgh. The total energy of the system is now that potential energy and must be the same as the total energy initially.

 

[dandirom]

ok, thank you very much. I'll try it out.

 

[dandirom]

hi there,

i tried this formula.

2 + 0.01
-------- sqr rt 2 * (9.8 m/s2) * 0.08m
0.01


i would appreciate your thoughts and comments.

 

[HallsofIvy]

hi there,

i tried this formula.

2 + 0.01
-------- sqr rt 2 * (9.8 m/s2) * 0.08m
0.01


i would appreciate your thoughts and comments.

My first thought is: "That's not a formula!" It's not equal to anything so it's not an equation.

If a mass m rises to height h then its potential energy has increased by mgh. Since the bullet and pendulum combined have mass 2.01 their potential energy increases by (2.01)(9.81)(0.08). Since the combined masses stop at that point they have 0 kinetic energy and so that is their combined energy. Initially, the bullet has speed v so it's kinetic energy is (1/2)(0.01)v². The pendulum is motionless so it's kinetic energy is 0 and both pendulum and bullet are at their lowest point so they have 0 potential energy. Initially, the total energy is (1/2)(0.01)v². Since total energy is conserved, (1/2)(0.01)v²= (2.01)(9.81)(0.08).

 

[dandirom]

My first thought is: "That's not a formula!" It's not equal to anything so it's not an equation.

If a mass m rises to height h then its potential energy has increased by mgh. Since the bullet and pendulum combined have mass 2.01 their potential energy increases by (2.01)(9.81)(0.08). Since the combined masses stop at that point they have 0 kinetic energy and so that is their combined energy. Initially, the bullet has speed v so it's kinetic energy is (1/2)(0.01)v². The pendulum is motionless so it's kinetic energy is 0 and both pendulum and bullet are at their lowest point so they have 0 potential energy. Initially, the total energy is (1/2)(0.01)v². Since total energy is conserved, (1/2)(0.01)v²= (2.01)(9.81)(0.08).

What about this?



sqrt (2)(9.8)(0.08)=1.25

2+0.01/0.01 * 1.25 = 250

 

[Zell2]

Initially, the total energy is (1/2)(0.01)v2. Since total energy is conserved, (1/2)(0.01)v2= (2.01)(9.81)(0.08).

Is this is right?

Since the bullet and the pendulum remain together, I'd have thought the kinetic energy of the bullet/pendulum together would only be a fraction of the kinetic energy of the bullet (the rest being dissipated as heat).

If the original poster knows that the kinetic energy of two particles can be split up into energy due to motion relative to the centre of mass and energy due to motion of the centre of mass I would use this. Otherwise work it out using conservation of momentum.

 

[Doc Al]

Since total energy is conserved, (1/2)(0.01)v²= (2.01)(9.81)(0.08).

Careful here. During the collision of bullet and block, as Zell2 points out, mechanical energy is not conserved.

Solve this in two steps, as mbrmbrg stated earlier:
(1) The collision of bullet with block. Momentum is conserved, but not mechanical energy.
(2) The rise of "bullet + block" after the collision. After the collision, energy is conserved.

 

[turdferguson]

Careful here. During the collision of bullet and block, as Zell2 points out, mechanical energy is not conserved.

Solve this in two steps, as mbrmbrg stated earlier:
(1) The collision of bullet with block. Momentum is conserved, but not mechanical energy.
(2) The rise of "bullet + block" after the collision. After the collision, energy is conserved.

And the trick is to work backwards from 2 to 1. With the height you can find the kinetic energy after the collision. Then you can determine the velocity of the block/bullet after the collision and finally the speed of the bullet before the collision using the inelastic collision formula

 

[Franksaei]

This is a method in the field to measure the initial velocity of a firearm. When the bullet exits from the barrel, it has a kinetic energy of E=1/2 x m x V^2. As the bullet hits the ball, the whole kinetic energy becomes the potential energy causing the ball rise to a certain elevation, which is E=Hmg where H is height, m is the mass of the ball and the bullet. Thus, to energies are equal and hens,
1/2 m v^2 = hmg
0.5 x 0.01 x v^2 = 0.08 x(2.00 + 0.01) x 9.81
and V^2= 315.168 v=17.75 m/s

In the real world of rifles, however, the pendulum goes higher than 8 cm and the material for pendulum has to be thick enough to keep the bullet inside and it has to be some kind of a hard and heavy wood

 

[Doc Al]

This is a method in the field to measure the initial velocity of a firearm. When the bullet exits from the barrel, it has a kinetic energy of E=1/2 x m x V^2. As the bullet hits the ball, the whole kinetic energy becomes the potential energy causing the ball rise to a certain elevation, which is E=Hmg where H is height, m is the mass of the ball and the bullet. Thus, to energies are equal and hens,
1/2 m v^2 = hmg
0.5 x 0.01 x v^2 = 0.08 x(2.00 + 0.01) x 9.81
and V^2= 315.168 v=17.75 m/s

No, energy is not conserved during the collision. Your formula will give too low a speed for the bullet.

(This thread is over 4 years old!)