# Linear momentum and Collisions

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.
Or vf =(m1-m2/m1+m2)*vi
Vf = (.1-1.2/.1+1.2)(6)
Vf= -5.076m/s

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

Can someone check my work?
Im not too sure about part b and c.
I think i used the wrong equations..

Last edited: Nov 14, 2009
2. Nov 15, 2009

### SeanGillespie

I think you may be wrong in part a).

My assumption is that part a) is based solely on the conservation of momentum.

The total momentum before a collision must equal the momentum after a collision:

[ (mass1 * velocity1) + (mass2 * velocity2) = Total momentum before ]

The second body (the spring) has no initial velocity, so the total momentum is the momentum at which the 0.1kg "shot" is moving.

After the collision, the bodies stick together, the product of their combined mass and their velocity should be equal to the momentum before the collision.

[Total momentum before = (mass1 + mass2) * velocity ]

As this velocity is your unknown variable, rearrange the formula to find it.

-------
Does anyone disagree with the above?

3. Nov 15, 2009

### willem2

a is wrong. Conservation of momentum implies:

$$(m_1 + m_2) v_f = m_1 v_1 + m_2 v_2$$

where $v_1$ and $v_2$ are the initial velocities and $v_f$ is the final velocity of both objects. You calculate this correctly in d. (the answer to d is the same as a)

For b find out the loss of kinetic energy of the objects involved in the collision. Since energy is conserved that loss must be stored in the spring.

c should be easy if you have b.

4. Nov 15, 2009

### JJBrian

I dont see how the answer of a is d.
I thought part a is an elastic collision while part d is an perfectly inelastic collision.
According to the formula used in part and d, they are both equal.
Can some explain this to me? I have a hard time visualizing this...
Some for part a)
(0.1kg + 1.2kg)*vf = (0.1kg)(6m/s)+(1.2kg)(0m/s)
Vf = 0.6/1.3 = 0.4615m/s

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

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

For 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) = 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??????
Im not too sure if my inputs for KE after collision are correct

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

### JJBrian

I need help with part b)

6. Nov 16, 2009

### willem2

you already solved b while solving c.

you use KE before collision = KE after collision + spring energyy and
Us = 1/2kx^2 to get x in one step, but you were supposed to get
the spring energy for question b and use that to get the compression of the spring
for question c.