Collision of Objects Dropped from Different Heights with Spring Compression

In summary: Just put the equation for the first object into the equation for the second object and you will get the position where they meet.Hope it helps, and please correct me if I was wrong anywhere.
  • #1
teme92
185
2

Homework Statement



(a) A block of mass m = 2.00 kg sits on a massless platform supported by a vertical
spring with spring constant k = 55.0 N/m. How much is the spring compressed?

(b) The block is now lifted off the platform and dropped from a height h = 1.5 m
above the platform and spring. What is the speed of the block just as it reaches
the spring?

(c) What is the maximum compression of the spring by the block?

(d) Just as the block leaves the spring moving upward, another identical block is
dropped from a height h2 = 2.5 m above the equilibrium position of the spring.
How far above the spring are the two blocks when they collide?


Homework Equations



F=ma
F=-kx
v2=u2 + 2as
KE=0.5mv2
PE=0.5kx2
W=0.5kd2

The Attempt at a Solution



(a)
I used ma=-kx and got 2(-9.8)=-55x →x=0.356m

(b)
Using v2=u2 + 2as, where s=-1.5, u=0, a=-9.8

I got v=5.422ms-1

(c)
I used W=KE + PE

And got d = 1.0935m

(d)
It is here where I have trouble. Could anyone tell me if I have used the right methods up til now? I get for part (d) I'm trying to find when both masses have equal h values but I don't know how to show that.
 
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  • #2
The potential energy ## U= \frac{1}{2} kx^2 ## in the spring is now converted to kinetic energy ## K = \frac{1}{2}mv^2 ##

Let ## s_1 ## represent the height of the block just about to leaving the spring, and ## s_2 ## be the height of the second identical block being dropped from rest.

Thus at any time t,

## s_1 = v_0 t - \frac{1}{2}gt^2 ##

Block on spring can be set at height = 0 (reference point). You have already solved ## v_0 = 5.6 m/s ## using mechanical conservation of energy. Since the block is thrown up it will start to decelerate immediately due to gravity.

At any time t the position of the second block is given by,

## s_2 = h - \frac{1}{2}gt^2 ##

Because the block starts at height ## s_2 = h_2 ## just sub that it for ##s_2##. We know the initial velocity is 0. The block once released is immediately accelerated downwards.

At the collision point the position of both blocks are the same i.e ## s_1 = s_2 ##

solve for t and sub this back into either ## y ## equation.
 
  • #3
Everything you've done up to the last part is correct.

The ending can be viewed in a couple ways, but the way that I think is probably the least work intensive is to look at the relative velocities of these objects.

If you were to compare the relative velocities, you could simplify this problem into one block going at a single velocity toward a static point. The relative velocity of the two objects would just be the difference of the two velocities. Well, since you know that they both have a constant downward acceleration, the acceleration aspect would end up cancelling out. The second block was dropped from rest, so the system can be simplified down to the single block's initial velocity going upward, and the time it down take for it to travel a distance of 2.5 meters.

Once you have this time, plug it back into either one of the motion equations for your original system of two blocks, and this will give you the distance at which they meet.
 
  • #4
Right so using the first equation I got:

0 = 5.6t - 0.5(-9.8)t2

Getting t on its own for t=1.1s

I then subbed that into the second equation for:

2.5 = h - 0.5(-9.8)(1.12)

Getting h on its own for h=3.429m which doesn't make sense so I must be going wrong somewhere. Thanks for the help by the way.
 
  • #5
teme92 said:
Right so using the first equation I got:

0 = 5.6t - 0.5(-9.8)t2

Getting t on its own for t=1.1s

I then subbed that into the second equation for:

2.5 = h - 0.5(-9.8)(1.12)

Getting h on its own for h=3.429m which doesn't make sense so I must be going wrong somewhere. Thanks for the help by the way.

Why did you set that to zero? That means you're solving for the time it would take to travel 0 meters, which doesn't make a ton of sense.
 
  • #6
I said the block was at height h = 0 not ##s_1 = 0 ## recall that ## y = x_0 + v_0t + \frac{1}{2} at^2 ## so ## x_0 = 0 ## for block on spring.
 
  • #7
## y_1 = v_0t - \frac{1}{2}gt^2 = h-\frac{1}{2}gt^2 = y_2 ##

## t = \frac{h}{v_0} ##

## y_1 = h - \frac{1}{2}g \Big(\frac{h}{v_0}\Big)^2 ##

## h = 2.5 m ##
 
  • #8
I have just studied the whole thing of springs and Energy few hours ago. But I thought I could share my idea.

I have got the velocity that the object on the string will launch off towards the second object. After all, The work that the spring does on the object will be equal to its kinetic energy.. I guess what you have a problem in is determining where they will meet.

Take a simple situation, two objects with a constant speed of v1 and v2.
The first object equation is: df = di + v1*t
Lets assume that di = 0
and the second object that is falling from a height let's just say h:
df = h - v2*t
Look at the minus there because it is going in the opposite direction.
What I want is the time that the two objects will collide on. You just have to put 1 into 2 and you will get your answer.
The same thing applies to the spring situation but you don't need to put the acceleration as rellek said you will just end up removing it.
 

1. What is mass and spring compression?

Mass and spring compression is a phenomenon that occurs when a mass is attached to the end of a spring and the spring is compressed or stretched. This causes the potential energy of the spring to increase, and the mass to experience a force in the opposite direction of the compression or stretching.

2. How is the amount of spring compression related to the mass?

The amount of spring compression is directly proportional to the mass attached to the spring. This means that as the mass increases, the amount of compression or stretching of the spring also increases.

3. What factors affect the amount of compression in a mass-spring system?

The amount of compression in a mass-spring system is affected by the stiffness of the spring, the mass attached to the spring, and the force applied to the spring.

4. What is the formula for calculating the amount of spring compression?

The amount of spring compression can be calculated using the formula F = -kx, where F is the force applied to the spring, k is the spring constant, and x is the amount of compression or stretching of the spring.

5. How does the motion of the mass in a mass-spring system change as the spring is compressed?

As the spring is compressed, the potential energy of the system increases and the mass will experience a force in the opposite direction. This will cause the mass to oscillate back and forth around its equilibrium position until the spring reaches its maximum compression. The motion of the mass will also depend on the initial conditions, such as the initial displacement and velocity of the mass.

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