An Object coliding with a spring gun

[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)?

Homework Equations

Anything that has to do with physics
Finial velocity
Potential Energy
Spring
ELASTIC AND Inelastic collisions

The Attempt at a Solution

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

 

[Jhero]

Hi there,

Thank you for your post and for showing your attempt at solving the problem. Your approach is mostly correct, but there are a few things that can be improved upon.

a) To find the recoil speed of the spring gun, we can use the conservation of momentum equation:

m1v1 + m2v2 = m1v1' + m2v2'

where m1 and v1 are the mass and initial velocity of the 0.1kg mass, m2 and v2 are the mass and initial velocity of the spring gun, and v1' and v2' are the final velocities of the two objects after the collision.

Since the spring gun is initially at rest, v2 = 0. And since the 0.1kg mass is lodged in the spring gun after the collision, we can assume that they move together with the same final velocity. So we have:

(0.1kg)(6m/s) = (0.1kg + 1.2kg)v'

v' = (0.1kg)(6m/s) / (0.1kg + 1.2kg)

v' = 0.4615m/s

b) The energy stored in the spring gun after the collision is equal to the potential energy stored in the compressed spring. So we can use the same equation that you used:

Us = 1/2kx^2

But we need to use the maximum compression distance for x, which we can calculate using Hooke's law:

F = -kx

F = (0.1kg)(9.8m/s^2)

x = (0.1kg)(9.8m/s^2) / (0.4N/m)

x = 2.45m

So the energy stored in the spring gun is:

Us = 1/2(0.4N/m)(2.45m)^2

Us = 0.6J

c) You have correctly calculated the maximum compression distance in part b, which is 2.45m. But for part c, we need to find the difference between this maximum compression and the relaxed position of the spring, which is 0. So the spring is compressed by 2.45m.

d) Your approach for part d is correct, but there is a small error in your calculation. It should be:

Vf = (m1