Elastic collision and inelastic collision

andylie

Hi everyone, im stuck with this problems.

1. 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)

2. 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

3. 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 cant 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

Related Introductory Physics Homework Help News on Phys.org

vela

Staff Emeritus
Homework Helper
Hi everyone, im stuck with this problems.

1. 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)

2. 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

3. 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 cant 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

andylie

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

vela

Staff Emeritus
Homework Helper
The spring force is conservative. What does that imply?

andylie

Is my answer correct? i plug in 99.99% of the energy is stored in the spring but its wrong.
I dont 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.

vela

Staff Emeritus
Homework Helper
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.

"Elastic collision and inelastic collision"

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