How can collisions and energy be related in a horizontal table scenario?

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The discussion focuses on a physics problem involving a 0.1 kg mass colliding with a 1.2 kg spring gun on a horizontal table. Key calculations include determining the recoil speed of the spring gun after maximum compression, which involves using energy conservation principles. The energy stored in the spring is calculated to be 1.8 J, while the maximum compression of the spring is found to be 3 m. Additionally, if the mass collides with a block of putty instead, the recoil speed is calculated to be approximately 0.4615 m/s. The discussion emphasizes the importance of following a systematic approach to solving the problem.
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Homework Statement


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)?
 
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Attempt ... I think my approach is totally wrong
I started with part c first...
I think the question should be answer in order.
The problem is I don't know how to approach question a with the given variables.
a)
Ws = (1/2)kx^2max
Ws = (0.4N/m)(3m)^2
Ws = 1.8J
Ws = 1/2mvf^2-1/2mvi^2
vf=sqrt(vi^2 +(2/m)*Ws
vf=sqrt(6m/s^2 +(2/.1kg)*(1.8J)
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
 
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