- #1
kky1638
- 4
- 0
I think I solved the problem but tell me if I made any errors
A tray of mass m, when suspended from a spring attached to the ceiling, stretches the spring by a distance Δx. A glob of peanut butter with mass M is dropped from a height h onto the tray, which is at rest, and sticks there.
a) What is the spring constant k of the spring in terms of the variables given above?
b) Find the maximum extension of the spring below its initial equilibrium position before the tray had been attached.
x = Acos(wt + @)
v = -Awsin(wt + @)
w = sqrt (k/mass) etc.
(a) Since the spring stretches Δx when tray is attached,
kΔx = mg, k = Δx/g
(b) The butter of mass M drops height h, so the velocity just before it sticks to the tray is
sqrt(2gh). And using conservation of momentum during the very short time period it sticks to the tray, M * sqrt(2gh) = (M + m)Vi, Vi = (M*sqrt(2gh))/(M + m)
Vi can be found from this and it is the v at time 0.
v(0) = -Awsin(@)
0 = x(0) = Acos(@)
A = sqrt((v(0)/-w)^2 + 0^2) = v(0)/w = Vi/w
So the maximum extension is (M*sqrt(2gh))/(M + m) / (sqrt(k/(M+m))) + Δx
Is this correct?
Is there a faster way to do this?
Homework Statement
A tray of mass m, when suspended from a spring attached to the ceiling, stretches the spring by a distance Δx. A glob of peanut butter with mass M is dropped from a height h onto the tray, which is at rest, and sticks there.
a) What is the spring constant k of the spring in terms of the variables given above?
b) Find the maximum extension of the spring below its initial equilibrium position before the tray had been attached.
Homework Equations
x = Acos(wt + @)
v = -Awsin(wt + @)
w = sqrt (k/mass) etc.
The Attempt at a Solution
(a) Since the spring stretches Δx when tray is attached,
kΔx = mg, k = Δx/g
(b) The butter of mass M drops height h, so the velocity just before it sticks to the tray is
sqrt(2gh). And using conservation of momentum during the very short time period it sticks to the tray, M * sqrt(2gh) = (M + m)Vi, Vi = (M*sqrt(2gh))/(M + m)
Vi can be found from this and it is the v at time 0.
v(0) = -Awsin(@)
0 = x(0) = Acos(@)
A = sqrt((v(0)/-w)^2 + 0^2) = v(0)/w = Vi/w
So the maximum extension is (M*sqrt(2gh))/(M + m) / (sqrt(k/(M+m))) + Δx
Is this correct?
Is there a faster way to do this?