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**[SOLVED] Inelastic collision on a frictional surface**

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- Thread starter aesnix
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In summary: Find the distance D the block moved on the surface:1/2(m+M)v'^2 - u(m+M)gD = 0D = [1/2(m+M)v'^2]/[u(m+M)g] = 1/2v'^2/ug.= 1/2v'^2/(ug)2. Find the distance d the bullet moved in the block:1/2mv^2 - fd = 1/2(m+M)v'^21/2mv^2 - fd = 1/2(m+M)(mv/(M+m))^21/2mv^2 - 2fd = 1/2

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A wood block with mass

Work = Force x Distance (constant parallel force)

inelastic collision: m1 x v0 = (m1 + m2) x v

I assumed that the collision was fast enough that the block didn't move apprecialbly during the collision, thus:

mv = (M+m)v'....so v'=mv/(M+m)

and, with energetic considerations:

for bullet m: 1/2mv^2 - fd = 1/2mv'^2

for block M: 1/2Mv'^2 - u(M+m)gD = 0

I know my assumption may well be wrong, but I had no clue of doing it in another way. I didn't include any heat that was lost either, but I don't know how. I guess I don't understand the characteristics of frictional force very well. The wiki says the heat lost during a movement on a frictional surface is f x d... uh, why?

According to Newton's 3rd law, I thought that if the surface doesn't move, then the work that was "supposed to done by an object on the surface" is lost as heat... but what if "the surface," as in the above case, IS moving? Also, if the Earth served as "the surface", wouldn't it acquire a very very small acceleration when an object slides on it?

Why, then. does the energy become heat but not the Earth's kinetic energy?

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aesnix said:Sorry I clcked the wrong button and sent it out before typing anything... I didn't even catch up on the 30-min editting time. Please forgive my slow typing...

## Homework Statement

A wood block with massMwas laid on a frictional surface with coefficient of kinetic frictionu. A bullet with massmflew towards the wook block with constant speedv; it shot into the block and moved along with the block for a distance before coming to rest. Given the force that the bullet experienced in the block wasf, try to find out the distance D the block moved on the surface, and the distance d the bullet moved in the block.

## Homework Equations

Work = Force x Distance (constant parallel force)

inelastic collision: m1 x v0 = (m1 + m2) x v

## The Attempt at a Solution

I assumed that the collision was fast enough that the block didn't move apprecialbly during the collision, thus:

mv = (M+m)v'....so v'=mv/(M+m)

and, with energetic considerations:

for bullet m: 1/2mv^2 - fd = 1/2mv'^2

for block M: 1/2Mv'^2 - u(M+m)gD = 0

Since the bullet and block are moving together, I think this needs to be:

[tex]

\frac{1}{2}(M+m) v'^2 - u (M+m) g D =0

[/tex]

I know my assumption may well be wrong, but I had no clue of doing it in another way. I didn't include any heat that was lost either, but I don't know how. I guess I don't understand the characteristics of frictional force very well. The wiki says the heat lost during a movement on a frictional surface is f x d... uh, why?

You did include the heat lost due to friction in your energy equation--that's the [itex]u (M+m) g D[/itex] term. The frictional force here is [itex]f_k=u(M+m) g[/itex].

According to Newton's 3rd law, I thought that if the surface doesn't move, then the work that was "supposed to done by an object on the surface" is lost as heat... but what if "the surface," as in the above case, IS moving? Also, if the Earth served as "the surface", wouldn't it acquire a very very small acceleration when an object slides on it?

Why, then. does the energy become heat but not the Earth's kinetic energy?

Any transfer to the Earth's rotational energy from this process would be so small that the distance traveled D would be the same whether you tried to include this effect or not.

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- #5

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aesnix said:According to Newton's 3rd law, I thought that if the surface doesn't move, then the work that was "supposed to done by an object on the surface" is lost as heat... but what if "the surface," as in the above case, IS moving? Also, if the Earth served as "the surface", wouldn't it acquire a very very small acceleration when an object slides on it?

Why, then. does the energy become heat but not the Earth's kinetic energy?

Hi aesnix! Welcome to PF!

(Actually, Newton's 3rd law is the one about equal and opposite reactions, which doesn't apply to heat! This is just conservation of energy …

http://en.wikipedia.org/wiki/Newton's_third_law#Relationship_to_the_conservation_laws:

Conservation of energy was discovered nearly two centuries after Newton's lifetime, the long delay occurring because of the difficulty in understanding the role of microscopic and invisible forms of energy such as heat and infrared light.

And in exam questions, you can

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Thanks for your help Tiny-tim.

Still, to avoid some misunderstanding arising from my poor statement, I think I should post the standard solution: (*the solution mentioned "internal energy," but I don't want to use this concept because it wasn't taught in class, nor was it mentioned in my text. Well this is what education is like in my country. :(* )

**mv = (M+m)v'...v' = mv/(M+m)**

1/2mv^2 - fd = 1/2(M+m)v'^2

1/2(M+m)v'^2 - u(M+m)gD = 0

__This is how I tried to interpret the sol in my way:__

I split the whole process into 2 stages-

1. From the instant when the bullet hit the block, the f had been doing negative work on the bullet, with the reaction force -f doing positive work on the block, until the bullet and the block had the same velocity.*Since f was a lot larger than u(M+m)g and the stage lasted a very short time, this stage can be treated as an inelastic collision with no external force applying on the system. ***(I doubt the validity of this assumption)** That is,

**mv = (M+m)v'**...(treat this stage as an inelastic collision with conserved momentum)

**1/2mv^2 - fd = 1/2(M+m)v'^2**...(a portion fd of the total kinetic energy was lost as heat, with the rest being the total kinetic energy at the end of stage 1)

2. Now the bullet and the block had the same velocity and could be seen as a composite body with mass M+m, speed v', moving on the rough surface. The frictional force from the surface did negative work on the body until it came to a stop. That is,

**1/2(M+m)v'^2 - u(M+m)gD = 0**...(the total kinetic energy was lost as heat)

Is that reasonable?

Still, to avoid some misunderstanding arising from my poor statement, I think I should post the standard solution: (

1/2mv^2 - fd = 1/2(M+m)v'^2

1/2(M+m)v'^2 - u(M+m)gD = 0

I split the whole process into 2 stages-

1. From the instant when the bullet hit the block, the f had been doing negative work on the bullet, with the reaction force -f doing positive work on the block, until the bullet and the block had the same velocity.

2. Now the bullet and the block had the same velocity and could be seen as a composite body with mass M+m, speed v', moving on the rough surface. The frictional force from the surface did negative work on the body until it came to a stop. That is,

Is that reasonable?

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aesnix said:Since f was a lot larger than u(M+m)g and the stage lasted a very short time, this stage can be treated as an inelastic collision with no external force applying on the system. (I doubt the validity of this assumption)

No, that's fine … momentum is ALWAYS conserved in collisions (whether elastic or not).

Is that reasonable?

Yes, that's very good!

oh … but

(Do you bother about where the

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No, that's fine … momentum is ALWAYS conserved in collisions (whether elastic or not).

Well what I meant was, can that stage be seen as a collision? More specifically, can I split the process into

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aesnix said:Well what I meant was, can that stage be seen as a collision? More specifically, can I split the process intostages? After all there was indeed a frictional force applied by the surface; that was anexternalforce for my system. Besides, the original question did not state that f was large.

Yes …

Bullets move

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Thanks a lot Tiny-tim!

An inelastic collision on a frictional surface is a type of collision where two objects come into contact and stick together after the collision. This means that the total kinetic energy of the system is not conserved, as some of it is lost due to friction between the two objects.

The outcome of an inelastic collision on a frictional surface can be affected by the mass, velocity, and coefficient of friction of the objects involved. The angle of collision and the surface properties of the objects and the surface they collide on can also play a role.

The coefficient of restitution is a measure of the elasticity of a collision. In an inelastic collision on a frictional surface, the coefficient of restitution will be less than 1, indicating that some of the kinetic energy is lost due to friction.

In an elastic collision on a frictional surface, the objects bounce off each other after the collision and the total kinetic energy is conserved. In an inelastic collision, the objects stick together and the total kinetic energy is not conserved due to energy loss from friction.

The outcome of an inelastic collision on a frictional surface can be calculated using the laws of conservation of momentum and energy. These equations take into account the mass, velocity, and coefficients of friction of the objects involved, as well as the angle of collision and the surface properties.

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