Can Physics Confirm the Height of a Carnival Ride Start?

[PascalPanther]

Homework Statement

Oscillations: You and some friends are waiting in line for "The Mixer", a new carnival ride. The ride begins with the car and rider (150 kg combined) at the top of a curved track. At the bottom of the track is a 50 kg block of cushioned material which is attached to a horizontal spring whose other end is fixed in concrete. The car slides down the track ending up moving horizontally when it crashes into the cushioned block, sticks to it, and oscillates at 3 repetitions in about 10 seconds. Your friends estimate that the car starts from a height of around 10 feet. You decide to use your physics knowledge to see if they are right. After the collision, you notice that the spring compresses about 15 ft from equilibrium.

Homework Equations

E = 1/2mv^2 + 1/2kx^2 = 1/2kA^2
omega = Sqrt(k/m)
f = omega/(2pi) = 1/(2pi)*Sqrt(k/m)

The Attempt at a Solution

m = 150kg
M = 150kg + 50kg = 200kg
x = 15ft = 4.57m
h1 = 3.04 m
h_real = ?
x = A?

mgh = (1/2)Mv^2 + (1/2)kx^2
Since there is no velocity given, I use:
mgh = (1/2)kA^2

I need to find k:
3 cycles every 10 seconds
f = 0.3 cycles/s

f = (1/2pi)Sqrt(k/M)
0.3 = (1/2pi)Sqrt(k/M)
k = 710 N/m

(150kg)(9.8m/s^2)(h) = (1/2)(710N/m)(4.57m)^2
h = 5.04m

h_real > h1

Did I do that correctly?

 

[Aero]

Almost: you have a rounding error in the final answer.

 

[Doc Al]

mgh = (1/2)Mv^2 + (1/2)kx^2
Since there is no velocity given, I use:
mgh = (1/2)kA^2

You have assumed that mechanical energy is conserved during the collision. Rethink that assumption, given that the car and block stick together.

I need to find k:
3 cycles every 10 seconds
f = 0.3 cycles/s

f = (1/2pi)Sqrt(k/M)
0.3 = (1/2pi)Sqrt(k/M)
k = 710 N/m

Good.

 

[PascalPanther]

Hmm...
Before the collision, the block would be:
mgh = (1/2)mv^2
Then
mv = Mv_2
Is there a way to use this relationship without velocity data (actual height)?

 

[Doc Al]

Sure. Start with the SHM. That should enable you to find the post-collision speed, then use that relationship (conservation of momentum) to find the pre-collision speed. Then you can deduce the height.

 

[PascalPanther]

(1/2)Mv^2 = (1/2)kA^2 right?

v = Sqrt(k/M)A
v = Sqrt(710/200kg) * (4.57m)
v = 8.6 m/s

mv = Mv
(150kg)v = (200kg)(8.6 m/s)
v= 11.5 m/s

mgh = (1/2)mv^2
(150kg)(9.8)h = (1/2)(150kg)(11.5m/s)^2
h = 6.75 m?

 

[Doc Al]

Looks good.