Physics Concept Question: Regarding springs and velocity/height of projectile

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

ghostman

Full Member
15+ Year Member
Joined
Aug 8, 2008
Messages
296
Reaction score
8
This isn't from any particular study guide. It's just a concept I was reviewing and wanted to make sure I had correct.

Object H has large mass (heavy).
Object L has small mass (light).

If H was placed on a Hooke's Law spring compressed by distance X and released such that H is projected straight up, the resulting initial velocity would be smaller than that of L in the same situation.
(1/2)kx^2 = (1/2)mv^2 :: solve for v

As a result, the max height for H will be less than L.
(1/2)kx^2 = mgh :: solve for h
or
v = sqrt(2gh) :: solve for h

If both H and L were placed on the spring AT THE SAME TIME and compressed by X, the initial velocities of both objects are the same and H and L will reach the same height.

This make sense and I'm fairly confident it's correct, but I haven't touched Physics in a while. Thanks.
 
This isn't from any particular study guide. It's just a concept I was reviewing and wanted to make sure I had correct.

Object H has large mass (heavy).
Object L has small mass (light).

If H was placed on a Hooke's Law spring compressed by distance X and released such that H is projected straight up, the resulting initial velocity would be smaller than that of L in the same situation.
(1/2)kx^2 = (1/2)mv^2 :: solve for v

As a result, the max height for H will be less than L.
(1/2)kx^2 = mgh :: solve for h
or
v = sqrt(2gh) :: solve for h

If both H and L were placed on the spring AT THE SAME TIME and compressed by X, the initial velocities of both objects are the same and H and L will reach the same height.

This make sense and I'm fairly confident it's correct, but I haven't touched Physics in a while. Thanks.

Which answer is correct? The one that says H will be less than L or that they will have equal initial velocities and reach equal height?
 
If both H and L were placed on the spring AT THE SAME TIME and compressed by X, the initial velocities of both objects are the same and H and L will reach the same height.
1/2kx^2 = 1/2mv^2 ----(1)
1/2kx^2 = 1/2MV^2 ----(2)

now since k and x are the same in eq. 1 and 2 you can see that the velocities must be different to compensate for the extra mass...so I dont think the statement is true.
I was wrong the first time around...


But in this case I think it would be alright to assume 1/2kx^2 = 1/2(M1 + M2)V^2 {This is because until the objects are released they can be treated as a single object with masses M1 and M2...So yes the initial velocity will be the same }

1/2M1V1^2 = M1gh {M1's and M2's cancel out in both equations)
and
1/2M2V2^2 = M2gh

(V1^2)/2g = h
(V2^2)/2g = h

Height is independent of mass since the initial Velocities were same V1 = V2 your statement was correct !
 
Last edited:
Just fixed this..my first explanation was wrong I was making an invalid assumption that the two systems were independent ! :smack:
 
Top