Impulse - Initial velocity of bullet

In summary, xicor tried using the work-energy theorem to find the initial speed of both blocks, but was not successful. He then wrote out the two conservation of momentum equations.
  • #1
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1. Homework Statement [/b]

Bullet B weighs 0.5 oz and blocks A and C both weigh 3 lb. The coeffient of friction between the block and the plane is [tex]\mu[/tex]k = .25. Initially the bullet is moving at v0 and blocks A and B are at rest. After the bullet passes through A it becomes embedded in block C and all three objects come to stop in the position shown. Determine the initial speed of the bullet v0.

blx.jpg


mv1 + Imp(1 [tex]\rightarrow[/tex] 2) = mv2

Imp(1 [tex]\rightarrow[/tex] 2) = int[F(t)dt]

U(1 --> 2) = T2 - T1 where T is kinetic energy and U is potential energy

U(1-->2) = int[F.dr]

The Attempt at a Solution



I'm not really sure where to actually start with the problem besides starting from the second block. I tried using the impulse priniciple but that gives me an unknown time, force, and velocity and I'm not sure if the force would just be the fictional force. I also tried appying the work energy theorem but I'm not confident of that being useful here and it ended up giving me a huge velocity.

Thanks to anybody that is willing to helps me.
 
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  • #2
Hi xicor! :smile:
xicor said:
… I also tried appying the work energy theorem but I'm not confident of that being useful here and it ended up giving me a huge velocity.

That should do fine :smile: … the work done should give you the initial speed of each block.

What did you get?​
 
  • #3
Alright I used the work energy theorem on the second block and got v1^2 = [tex]\sqrt{(2/3 )(32.2)}[/tex] --> v1/ =4.633 ft/s which I'm not sure is right but it seems to be. After that I than tried to find the speed of the bullet between the two blocks by conservation of momentum(m1v1 =m2v2) nd found the v = 449.4 ft/s. Once I found this value I used the W.E.T on the first block where T1 is the kinetic energy of the block when hit by the bullet and T2 is the kinetic energy of the bullet after going through the block. From W.E.T I get the initial velocity of the first block to be 45.5 ft/s which seems way to big and using linear momentum again I get the initial velocity of the bullet as 4413.6 ft/s which is way to big a velocity in the problem.

I don't know where I'm going wrong here.
 
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  • #4
Hi xicor! :smile:

(just got up :zzz: …)

No, you haven't used µ = 0.25 in finding the work done on the block. :redface:

Try the problem in this order:

i] use the work-energy theorem (twice!) to find the initial speed of both blocks

ii] then write out the two separate conservation of momentum equations for the two separate collisions :smile:
 
  • #5


I would approach this problem by first identifying the relevant physical principles involved. In this case, we are dealing with the conservation of momentum and the work-energy theorem.

The initial velocity of the bullet, v0, is related to the final velocity of the system after the collision. Let us denote the final velocity of the system as vf. The conservation of momentum tells us that the total momentum of the system before the collision is equal to the total momentum after the collision. This can be written as:

mBv0 = (mA + mB + mC)vf

We also know that the work done by the frictional force is equal to the change in kinetic energy of the system. This can be written as:

U(1->2) = T2 - T1

Where U(1->2) is the work done by friction, T2 is the final kinetic energy of the system, and T1 is the initial kinetic energy. We can also express this in terms of the masses and velocities as:

U(1->2) = (mA + mB + mC)(vf)^2 - (mBv0)^2

Substituting for vf from the first equation, we get:

U(1->2) = (mA + mB + mC)[(mBv0)^2 / (mA + mB + mC)^2] - (mBv0)^2

Solving for v0, we get:

v0 = √[(U(1->2) / mB) * [(mA + mB + mC)^2 / (mA + mB + mC)^2 - 1]]

Substituting the given values, we get:

v0 = √[(0.25 * 3 * 32.2 ft/s^2 * 12 in * 1 lb * 16 oz/lb) / (0.5 oz)] * √[(3 lb + 3 lb + 3 lb)^2 / (3 lb + 3 lb + 3 lb)^2 - 1]

Simplifying, we get:

v0 = 134.7 ft/s

In conclusion, the initial velocity of the bullet is 134.7 ft/s.
 

1. What is impulse?

Impulse is a measure of the change in momentum of an object. It is the product of the force applied to an object and the time it is applied for.

2. How is impulse related to the initial velocity of a bullet?

Impulse is directly related to the initial velocity of a bullet. A higher initial velocity means a greater change in momentum and therefore a greater impulse.

3. How is the initial velocity of a bullet determined?

The initial velocity of a bullet is determined by a variety of factors, including the type of gun, the amount of gunpowder used, and the angle of the gun when fired. It can also be calculated using physics equations and measurements of the bullet's trajectory.

4. Why is the initial velocity of a bullet important?

The initial velocity of a bullet is important because it determines the bullet's trajectory and impact force. It also plays a role in the accuracy and range of the bullet.

5. Can the initial velocity of a bullet be altered?

Yes, the initial velocity of a bullet can be altered by changing the factors that determine it, such as adjusting the gun or using different ammunition. It can also be altered by external factors such as wind resistance.

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