Impulse - Initial velocity of bullet

[xicor]

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 $$\mu$$_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.

mv₁ + Imp(1 $$\rightarrow$$ 2) = mv₂

Imp(1 $$\rightarrow$$ 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.

 

[tiny-tim]

Hi xicor!

… 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 … the work done should give you the initial speed of each block.

What did you get?​

 

[xicor]

Alright I used the work energy theorem on the second block and got v₁^2 = $$\sqrt{(2/3 )(32.2)}$$ --> v_1/ =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.

 

[tiny-tim]

Hi xicor!

(just got up :zzz: …)

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

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

 

[DeeNos]

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.