# Finding the recoil speed, energy stored, and compression of a spring

1. Nov 14, 2009

### JJBrian

1. The problem statement, all variables and given/known data

A 0.1kg is shot with a speed of 6m/s toward a 1.2kg spring gun( with spring constant of 0.4N/m). The spring gun is initially at rest with its spring relaxed. The spring gun is free to slide without friction on a horizontal table. The 0.1 kg mass compresses the spring to its maximum and remains lodged at this maximum compression.

a)what is the recoil speed of the spring gun( with the 0.1kg mass) after this event?

b)What is the energy stored in the spring gun after this event?

c) How much is the spring compressed from its relaxed position?

d) If instead of hitting a spring gun, this 0.1kg mass hit a 1.2 block of putty ( and stuck to the putty) that was free to slide with no friction on a horizontal table, what would be the recoil speed of the putty( with the 0.1 kg mass)?

2. Relevant equations
Linear motion and its conservation
Collisions

3. The attempt at a solution

a)

---------------------------------------

m1v1f +m2v2f = 0
v1f = -(m2/m2)*v2f
V1f=(-0.1kg/1.2kg)(6m/s)
v1f = -0.5m/s
Im not too include the spring constant for this part.

b)
Us = 1/2kx^2
Us = 1/2(0.4N/m)(3m)
Us = 0.6J

c)KE + Us = KE+ Us
0 +1/2kx^2max = 1/2mv^2 + 0
xmax =sqrt(m/k)*V
xmax = sqrt(.1kg/.4N/m)*(6m/s)
xmax = 3m

d) Vf = (m1/m1+m2)*v0
Vf = (.1kg/.1kg+1,2kg)*(6m/s)
Vf = 0.4615m/s

Last edited: Nov 14, 2009
2. Nov 14, 2009

### Delphi51

Fascinating problem! Hats off to the prof who came up with this one.

Okay, starting with your part (c):
c)KE + Us = KE+ Us
0 +1/2kx^2max = 1/2mv^2 + 0
It seems to me this should read
KE before collision = KE after collision + spring energy
I guess you left out the KE after the collision.
Of course you can't find it until you do part (a).

momentum before = momentum after
calculation to get the speed of the mass+spring gun after the collision.

According to my calc, the KE after the collision is quite small so your answer for x isn't very much too large!

3. Nov 14, 2009

### JJBrian

I changed part a.
I think B and C are wrong.
I dont know how to approach B and C.
I need some ideas, like the eq to use.

4. Nov 15, 2009

### Delphi51

For part (a) you use
momentum before collision = momentum after collision
mv = mv
Before the collision only the 0.1 kg shot is moving. After, the shot+gun are moving together.

5. Nov 15, 2009

### JJBrian

Updated work
part a)
(m1+m2)vf = m1v1 + m2v2
(0.1kg + 1.2kg)*vf = (0.1kg)(6m/s)+(1.2kg)(0m/s)
Vf = 0.6/1.3 = 0.4615m/s

For part b)
Do I just use KE =1/2mv^2
1/2(.1kg)(6m/s)^2
KE = 1.8J

part c)

KE before collision = KE after collision + spring energy
0 +1/2mv^2 = 1/2mv^2 + 1/2kx^2
1/2(.1kg)(6m/s)^2 = 1/2(0.1+1.2kg)(0.4615m/s)^2+1/2(0.4N/m)x^2
1.8J = 0.1384 + .2x^2
x=2.882m??????

d)Vf = (m1/m1+m2)*v0
Vf = (.1kg/.1kg+1,2kg)*(6m/s)
Vf = 0.4615m/s

Last edited: Nov 15, 2009
6. Nov 15, 2009

### Delphi51

For part (b), just use E before = E after
After the collision you'll have kinetic and spring energy. The spring energy is not 1.8 J. Oh, just what you did for (c). I agree with your answer for (c), BTW.
And (d) has to be the same as (a), no calc needed.

7. Nov 15, 2009

### JJBrian

For part b) I thought the kinetic energy of the object would be converted to the energy stored of the spring.
Would the energy stored in the spring be 1.66153J?
E before = E after
1/2mv^2 = 1/2mv^2 + 1/2kx^2
1.8J = 0.1384 + 1.66153

8. Nov 15, 2009

### Delphi51

Yes, 1.66 is what I have.

9. Nov 16, 2009