How Does Friction Affect Spring Compression in a Bullet Collision?

[Deadawake]

Hi,
Please see the attach picture.
I solved 3 quetions regarding this problem and I struggle with the 4th one.
What I found was that the initial velocity after the collision is 1.98 m/s (Edit: 0.2m/s), and the maximum compression of the spring with no friction beneath is 0.44m (Edit: 0.044m) (I just used energy conservation law and compare the initial kinetic energy to the potential spring energy)
But then they added the friction thing...
The question they ask is what is the maximum spring compression after the collision if the coefficient of kinetic friction is 0.2

What I did:
x = Maximum spring compression
Calculated the force by of friction (Fr) : -9.898N (Edit: -9.81)
Calculated the Work done by friction : -9.89⋅x [J] (Edit: -9.81x)
Used the conservation law : ½κx² + Fr = ½mv²

In this point I thought I was doing wrong , because where is the energy of heat due to friction goes?
if I use the conservation law can I neglect the heat energy ? it's none conservative force and in the other hand the "question" doesn't mention any detail about that.

I continued to solve this equation to figure out if I get some realistic answer.
so: ½⋅100⋅x² + (9.81⋅x) = ½⋅5.005⋅(-0.2)²

x(1,2) = 0.35m , -0.55m (Edit: 0.0097m , -0.205m)

The ~1cm sounds logic. but I don't know for sure if I did it right.
If i did, what is the other number? (-0.205m)
Thanks a lot,
And I honor the people who read all this.

 

[TomHart]

Are you sure you got your velocity after collision right? It's possible I didn't.

 

[Deadawake]

Are you sure you got your velocity after collision right? It's possible I didn't.

I think so -
According to the momentum conservation principle:
m₁v₁+m₂v₂ = (m₁+m₂)V_final
V_f = 5⋅0+200⋅0.05 / 5.05 = 1.98 m/s

 

[TomHart]

The bullet weight is 5 g = 0.005 kg (not 0.05 kg).
Edit: Oops. I mean bullet mass, not weight.

 

[Deadawake]

The bullet weight is 5 g = 0.005 kg (not 0.05 kg).
Edit: Oops. I mean bullet mass, not weight.

Ops... :0
I'm correcting:
Compression with no friction 0.044m or 4.4cm
Velocity after collision: -0.2 m/s
F_k = -9.81N

½⋅100⋅x² + 9.81⋅x = ½⋅5.005⋅(-0.2)²

x_1,2 = 0.0097m (/1cm), -0.206m (/-20cm)

Still I left with the same question

 

[TomHart]

When I think about problems like this, I kind of think of it this way. The block/bullet combination has some initial kinetic energy. If there is no friction, obviously all of that energy goes into compressing the spring. If there is friction, some of that energy is used up in friction and gets converted to heat - which means the distance will be less.

So I would write: E_kinetic = E_spring + E_friction
All of those terms will be positive numbers. Positive numbers are easier than negative numbers for me. :)

I got the same answer as you - namely, 0.0097 m, or slightly less than 1 cm.
One thing you could do to try to see if your answer makes sense. Calculate how far the block would slide if there was no spring - just the block and friction.

The negative number in your answer is meaningless to me, although some of the smart folks on here could give you some interpretation of it.

 

[zwierz]

if the bullet has got stuck in the cube then energy conservation law does not suit even if the cube slides without friction

 

[TomHart]

if the bullet has got stuck in the cube then energy conservation law does not suit

In the collision between the bullet and block, you are correct; energy conservation does not apply. But he did not work the problem in that manner. He was applying conservation of energy after the collision.

 

[zwierz]

n the collision between the bullet and block, you are correct; energy conservation does not apply. But he did not work the problem in that manner. He was applying conservation of energy after the collision.

o sorry, I see

 

[TomHart]

@zwierz, if you were commenting on my post, after I re-read my post, I realize I should have made it clearer that my energy equation was for post-collision. Sorry about that.

 

[haruspex]

what is the other number? (-0.205m)

It is not unusual, particularly with nonlinear equations, that the equation you obtain is also the equation for a different question. In the present case, because you have to square the speed the equation does not care about the direction. Of course, if the direction were reversed then the friction would be too, but if we replace the stationary ground with a belt moving rapidly to the right then that preserves the friction direction, leading to the same equation.

 

[Deadawake]

Thank you .
So did I do it right or am I missing something?
Can someone explain me why in the moment of the collision (the impact) the conservation law does not apply?

 

[TomHart]

Can someone explain me why in the moment of the collision (the impact) the conservation law does not apply?

Whenever a collision occurs where 2 bodies stick together, that is a perfectly inelastic collision. For that type of collision, conservation of momentum does apply, but conservation of energy does not apply. (Actually, at the micro level, energy is conserved because some of the energy gets converted to heat. But for the purpose of calculating velocities of the bodies after the collision, energy is not conserved.) In fact, for every type of collision except a perfectly elastic collision, it is true that momentum is conserved but energy is not conserved. Only in a perfectly elastic collision are both momentum and energy conserved.

 

[haruspex]

Thank you .
So did I do it right or am I missing something?
Can someone explain me why in the moment of the collision (the impact) the conservation law does not apply?

Which conservation law?

Conservation of momentum of a system applies as long as there is no exchange of momentum with something outside of that system. In this problem, the impact between bullet and block is so brief that there is not time for any significant exchange of momentum between block and spring or between block and ground.

Conservation of work should not generally be assumed without good reason. If the bullet had passed right through the block you would not know how much work was lost (if any) without further information. But you can show that if no work were lost then either the bullet lost no speed and the block remained stationary or the bullet bounced back off the block elastically!
Here you know the bullet and block coalesce. That corresponds to the maximum possible loss of work, consistent with conservation of momentum.