How Does a Bullet's Initial Speed Relate to Energy Conservation in a Collision?

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The discussion focuses on the relationship between a bullet's initial speed and energy conservation during a collision with a wooden block. The bullet embeds itself in the block, compressing a spring while overcoming friction. Participants clarify that the kinetic energy of the bullet-block system is partially converted into elastic potential energy in the spring and lost to friction. The correct approach involves calculating the total kinetic energy at impact and accounting for energy lost to friction, which does not contribute to the spring's potential energy. Ultimately, understanding the energy transformations is crucial for solving the problem accurately.
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Homework Statement


A bullet of mass 1.4×10−3 {\rm kg} embeds itself in a wooden block with mass 0.999 {\rm kg}, which then compresses a spring (k = 110 {\rm N/m}) by a distance 5.5×10−2 {\rm m} before coming to rest. The coefficient of kinetic friction between the block and table is 0.46.
a)what is the initial speed of the bullet?
b)What fraction of the bullet's initial kinetic energy is dissipated (in damage to the wooden block, rising temperature, etc.) in the collision between the bullet and the block? (answer: ΔK/K


The Attempt at a Solution


i started by drawing a diagram labeling part A to be where the spring is compressed, part B to be where the block started pushing the spring, part C to be where the bullet impacts the block, and part D the firing of the bullet.
I then used energy conservation, stating that the E@A=E@B. 1/2kx^2=1/2mv^2, solving for v and getting sq.rt of (kx^2/m). plugging in the numbers i get v=.5767 m/s. Then i attempt to use energy conservation from B to C, 1/2mv^2=1/2mv^2-(work by friction+work by spring) but i kind of just put that equation together myself because i know energy is lost by work done) and i don't really know what to do.
 
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While the spring is being compressed, where is the KE of the block+bullet going? Is it only into the spring's PE?
 
gneill said:
While the spring is being compressed, where is the KE of the block+bullet going? Is it only into the spring's PE?

well there is also friction, so I am guessing some of the KE from the block+bullet is being lost due to work done by friction.
 
bfusco said:
well there is also friction, so I am guessing some of the KE from the block+bullet is being lost due to work done by friction.

Correct. So when you want to find the velocity of the block+bullet when it first encounters the spring, both energies need to be accounted for. How much energy should be attributed to friction while the spring compresses?
 
gneill said:
Correct. So when you want to find the velocity of the block+bullet when it first encounters the spring, both energies need to be accounted for. How much energy should be attributed to friction while the spring compresses?

um...the energy attributed to friction is equal to the work done by friction? so I am going to say FΔx, the force of friction is equal to mass of bullet+block x g x μ_k x distance spring compressed= .248 J
 
bfusco said:
um...the energy attributed to friction is equal to the work done by friction? so I am going to say FΔx, the force of friction is equal to mass of bullet+block x g x μ_k x distance spring compressed= .248 J

Correct again :smile:

So what's the total kinetic energy that the block+bullet had when it first hit the spring?
 
gneill said:
Correct again :smile:

So what's the total kinetic energy that the block+bullet had when it first hit the spring?

ok so the total kinetic energy that the block+bullet system has is equal to 1/2mv^2+W(spring)+W(friction). if that is correct i understand the equation, however I am not understanding exactly how that equation works. when i substitute the numbers into solve for the velocity the system has before it hits the spring i get the sq. rt of a negative answer. i understand the energy part of this question but I am not sure when it comes to actually applying the formula.

what i did was: (EPE=elastic potential energy) EPE_A=KE_B+Ws+Wf, that becomes 1/2kx^2=1/2mv^2+(.248+.333). (the .333 is the work done by the spring, the force of the spring is F=kx=6.05N, W=FΔx=(6.05)(.055)=.333.) substituting the values i get 1/2(110)(.055)^2=1/2(1.0004)v^2+.581, doing that i get [sq.rt -.4146/.5002] which can't be.
 
When the spring is fully compressed, the energy stored in it is what was left of the original KE minus what was lost due to friction:

KE = PEspring + Wfriction

Remember, the energy lost to friction while the spring was compressing never makes it into the spring's potential energy -- it was lost.
 
gneill said:
When the spring is fully compressed, the energy stored in it is what was left of the original KE minus what was lost due to friction:

KE = PEspring + Wfriction

Remember, the energy lost to friction while the spring was compressing never makes it into the spring's potential energy -- it was lost.

wow. i got it, thank you. so i kind of understood it incorrectly, the KE at time of impact is = to the energy now in the spring plus the work done by friction. the work done by the spring isn't important here because that energy is going into the elastic potential energy not being lost by work.
 
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