Elastic collision and inelastic collision

In summary, the conversation discusses a physics problem involving two blocks of different masses and a spring. The initial mechanical energy of the system is calculated using the equations for momentum and energy conservation. The maximum compression of the spring is also determined. The conversation then moves on to discuss the energy lost during an inelastic collision and the energy stored in the spring during an elastic collision. The correct equations to use for each scenario are explained.
  • #1
andylie
9
0
Hi everyone, I am stuck with this problems.

Homework Statement


A block (m1 = 2.5 kg) sits at rest on a horizontal frictionless surface, connected to an unstretched spring (k = 190 N/m) whose other end is fixed to a wall. Another block (m2 = 1.0 kg) whose speed is 4.0 m/s collides head-on with it.

http://www.webassign.net/hrw/10_44alt.gif

a.) If the two blocks stick together after the collision,

of the initial mechanical energy of the (m1, m2 & spring) system:
how much is lost to the environment during the collision? _____ %
how much is stored in the spring at maximum compression? _____ %

what is that maximum compression of the spring (when the blocks momentarily stop)?
0.1551 m (i got this one right)


b.) If the two blocks instead collide elastically,

what will be the speed and direction of each block immediately after the collision?
block 1: _____m/s,
block 2: _____m/s, left( travel to the left is the right answer)

of the initial mechanical energy of the system:
_____how much is lost to the environment during the collision? %
_____how much is stored in the spring at maximum compression? %
(Where is the rest of the energy?)

what will be the maximum compression of the spring afterwards?
_____m ( i put 0meters since this is elastic collision, because the spring will compressed and return to its initial state, but the answer is wrong)






Homework Equations


This is about momentum and conservation of energy

formula:
(i=initial. f=final)

momentum conservation
pi=pf
(m1v1)i+(m2v2)i=(m1v1)f+(m2v2)f

energy conservation (KE=kinetic energy, PE=potential energy)
Ei=Ef
KEi+PEi=KEf+PEf
(1/2mv^2)i+(mgh)i=(1/2mv^2)f+(mgh)f

The Attempt at a Solution



heres how i find the maximum compression of the spring

first i use momentum conservation equation

(m1v1)i+(m2v2)i=(m1v1)f+(m2v2)f
(2.5*0)+(1*4)=(2.5*vf)+(1*vf)
4=(2.5+1)vf
4/(2.5+1)=vf=1.1428m/s

now i need to find the maximum distance the spring compressed
so i already know the velocity of (m1+m2) when they start compress the spring.
i set up the equation as
KEblock=KEspring
1/2mv^2=1/2kx^2 (where k is spring constant of 190N, x is distance)
we are looking for x, so i rearrange the equation
((1/2mv^2)*(2/k))^(1/2)=x
((mv^2)/k)^(1/2)=x
((3.5*1.1428^2)/190)^(1.2)= x =0.1551m

how much is lost to the environment during the collision? _____ % (inelastic)
i didnt get this one right, but this is how i do
KEi=KEf
KE(block2)=KE(block1+2)+PEspring
(1/2m2v^2)=(1/2(m1+m2)v^2)+(1/2kx^2)
(1/2*1*4*2)=(1/2(2.5+1)*1.1428^2)+(1/2(190)*0.1551^2)
8=2.28548+2.2843
8=4.5707

to find percent lost
((KEfinal-KEinitial)/KEinitial)*100)=((4.5707-8)/8)*100)=42.86% ( either this or -42.86% is wrong)

I have tried 2-3 times on some problems but i can't get it right. If anyone can explain and show me what equation should i use for all of them, i am really grateful because i am a person that learn through visual not listen. Again, thank you for your help and i really appreciate it
 
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  • #2
andylie said:
Hi everyone, I am stuck with this problems.

Homework Statement


A block (m1 = 2.5 kg) sits at rest on a horizontal frictionless surface, connected to an unstretched spring (k = 190 N/m) whose other end is fixed to a wall. Another block (m2 = 1.0 kg) whose speed is 4.0 m/s collides head-on with it.

http://www.webassign.net/hrw/10_44alt.gif

a.) If the two blocks stick together after the collision,

of the initial mechanical energy of the (m1, m2 & spring) system:
how much is lost to the environment during the collision? _____ %
how much is stored in the spring at maximum compression? _____ %

what is that maximum compression of the spring (when the blocks momentarily stop)?
0.1551 m (i got this one right)


b.) If the two blocks instead collide elastically,

what will be the speed and direction of each block immediately after the collision?
block 1: _____m/s,
block 2: _____m/s, left( travel to the left is the right answer)

of the initial mechanical energy of the system:
_____how much is lost to the environment during the collision? %
_____how much is stored in the spring at maximum compression? %
(Where is the rest of the energy?)

what will be the maximum compression of the spring afterwards?
_____m ( i put 0meters since this is elastic collision, because the spring will compressed and return to its initial state, but the answer is wrong)






Homework Equations


This is about momentum and conservation of energy

formula:
(i=initial. f=final)

momentum conservation
pi=pf
(m1v1)i+(m2v2)i=(m1v1)f+(m2v2)f

energy conservation (KE=kinetic energy, PE=potential energy)
Ei=Ef
KEi+PEi=KEf+PEf
(1/2mv^2)i+(mgh)i=(1/2mv^2)f+(mgh)f

The Attempt at a Solution



heres how i find the maximum compression of the spring

first i use momentum conservation equation

(m1v1)i+(m2v2)i=(m1v1)f+(m2v2)f
(2.5*0)+(1*4)=(2.5*vf)+(1*vf)
4=(2.5+1)vf
4/(2.5+1)=vf=1.1428m/s

now i need to find the maximum distance the spring compressed
so i already know the velocity of (m1+m2) when they start compress the spring.
i set up the equation as
KEblock=KEspring
1/2mv^2=1/2kx^2 (where k is spring constant of 190N, x is distance)
we are looking for x, so i rearrange the equation
((1/2mv^2)*(2/k))^(1/2)=x
((mv^2)/k)^(1/2)=x
((3.5*1.1428^2)/190)^(1.2)= x =0.1551m

how much is lost to the environment during the collision? _____ % (inelastic)
i didnt get this one right, but this is how i do
KEi=KEf
KE(block2)=KE(block1+2)+PEspring
(1/2m2v^2)=(1/2(m1+m2)v^2)+(1/2kx^2)
(1/2*1*4*2)=(1/2(2.5+1)*1.1428^2)+(1/2(190)*0.1551^2)
8=2.28548+2.2843
8=4.5707
In an inelastic collision, the kinetic energy isn't conserved. You found the kinetic energy before the collision is 8.00 J. Immediately after the collision, the kinetic energy is 2.29 J. How much was lost?

The spring doesn't have kinetic energy. The energy it has is potential, so you shouldn't have the spring term in there. Also, right after the collision, when the masses move with speed vf=1.14 m/s, the spring isn't compressed, so it again doesn't make sense to add in the term you did, which is when the spring is compressed and the blocks are stopped.

to find percent lost
((KEfinal-KEinitial)/KEinitial)*100)=((4.5707-8)/8)*100)=42.86% ( either this or -42.86% is wrong)

I have tried 2-3 times on some problems but i can't get it right. If anyone can explain and show me what equation should i use for all of them, i am really grateful because i am a person that learn through visual not listen. Again, thank you for your help and i really appreciate it
 
  • #3
Thank you, i got that one correct now. how about this one?
how much is stored in the spring at maximum compression? ____%
Since the kinetic energy of the block will be stored in the spring until the spring reach max compression
so i set the equation as

KEblock=PEspring
1/2(m1+m2)v^2=1/2kx^2
1/2(3.5)(1.1428^2)=1/2(190)(0.1551^2)
2.28548=2.28532

percent difference is
((PEspring final - KEblock final)/KEblock final)*100
((2.28532-2.28548)/2.28532)*100= -.00700%

so total percentage of energy stored is 100-0.007=99.992%

is that correct? i already tried 4 times and 1 more time error i will lose the point. thanks
 
  • #4
The spring force is conservative. What does that imply?
 
  • #5
Is my answer correct? i plug in 99.99% of the energy is stored in the spring but its wrong.
I don't mean to be rude and i really appreciate your help if you can just tell me what i did is right or wrong and show me how to approach this problem instead of asking me back what am i thinking.
If i am smart and good in physics i wouldn't go around asking for help.
 
  • #6
No, your answer is wrong. This is a conceptual question. If you understand the concepts, the answer is obvious — no calculation needed. That's why I pointed out that the spring force is conservative.
 

1. What is an elastic collision?

An elastic collision is a type of collision between two objects where the total kinetic energy of the system is conserved. This means that the objects bounce off each other without any loss of energy and the total momentum of the system is also conserved.

2. What is an inelastic collision?

An inelastic collision is a type of collision between two objects where the total kinetic energy of the system is not conserved. This means that some energy is lost during the collision, usually in the form of heat or deformation of the objects.

3. How is the kinetic energy of a system calculated in an elastic collision?

In an elastic collision, the total kinetic energy of the system before and after the collision is the same. This can be calculated using the equation KE = 1/2 mv2, where m is the mass of the object and v is the velocity.

4. What is the coefficient of restitution in a collision?

The coefficient of restitution is a measure of the elasticity of a collision. It is defined as the ratio of the relative velocity of the two objects after the collision to the relative velocity before the collision. In an elastic collision, the coefficient of restitution is 1, while in an inelastic collision, it is less than 1.

5. Can an elastic collision occur in real life?

Yes, elastic collisions can occur in real life. However, they are rare as most collisions involve some amount of energy loss. Examples of elastic collisions in real life include collisions between atoms and subatomic particles.

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