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

[JJBrian]

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)?

 

[JJBrian]

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

 

[kerimek]

In this scenario, collisions and energy are related through the principles of conservation of momentum and conservation of energy. When the 0.1kg mass collides with the spring gun, the momentum of the system is conserved, meaning that the total momentum before and after the collision remains the same. This can be expressed as m1v1 + m2v2 = m1v1' + m2v2', where m1 and v1 are the mass and velocity of the 0.1kg mass before the collision, m2 and v2 are the mass and velocity of the spring gun before the collision, and v1' and v2' are the velocities of the 0.1kg mass and spring gun after the collision.

In terms of energy, the collision between the 0.1kg mass and the spring gun can be thought of as a transfer of kinetic energy to potential energy. As the 0.1kg mass compresses the spring, it stores potential energy in the spring. This can be calculated using the equation E = 1/2kx^2, where k is the spring constant and x is the displacement of the spring from its relaxed position.

a) The recoil speed of the spring gun can be calculated using the conservation of momentum equation mentioned above. Since the 0.1kg mass is lodged in the spring, its velocity after the collision will be zero. Thus, the equation becomes m1v1 + m2v2 = m2v2', where v2' is the recoil speed of the spring gun. Solving for v2', we get v2' = m1v1/m2 = (0.1kg)(6m/s)/(1.2kg) = 0.5m/s.

b) The energy stored in the spring gun after the collision can be calculated using the equation mentioned above. Plugging in the values, we get E = 1/2(0.4N/m)(x^2), where x is the maximum compression of the spring. Since the 0.1kg mass is lodged in the spring, the maximum compression will be equal to the displacement of the spring from its relaxed position. Thus, the energy stored in the spring gun will be equal to the potential energy stored in the spring, which can be calculated using the equation mentioned above.

c) The spring will be compressed by an amount equal to the displacement of the 0.