Putty dropped on a frame suspended from a spring.

[jax988]

Homework Statement

A 0.150 kg frame, when suspended from a coil spring, stretches the spring 0.050 m. A 0.200 kg lump of putty is dropped from rest onto the frame from a height of 30.0 cm

The collision between the two is perfectly inelastic.

Find the maximum distance the frame moves downward from its initial position.

Homework Equations

$$\Delta{K} + \Delta{U} + \Delta{U_o} = 0$$

The Attempt at a Solution

I plugged in the energies into the above equations, and it didn't bring me closer to an answer. I'm not sure how I can relate momentum to energy in this, with a completely inelastic collision.

Any help is appreciated. Thank you.

 

[supratim1]

hello Jax988! welcome to PF!

remember that momentum is conserved. so find the common velocity of putty ad frame.

then with this velocity as 'v', and m = mass of putty + frame, find kinetic energy (initial)

then equate it with 1/2.k.x^2 + mgx
k = spring constant.

x is the required distance.

 

[ashishsinghal]

I plugged in the energies into the above equations, and it didn't bring me closer to an answer. I'm not sure how I can relate momentum to energy in this, with a completely inelastic collision.

In inelastic collision mechanical energy is not conserved. So you cannot directly use energy conservation. But there is no harm in using it when collision is completed.

 

[jax988]

I guess my problem is the quadratic as well. If I carry out the energy equation after the collision, I have the distance x included. However, it takes the form of a quadratic equation.

When using the equation given.

$$\frac{1}{2}Mv_0^2 - Mgd + \frac{1}{2}kd^2 - \frac{1}{2}kx_f^2 = 0$$

Where $$M$$ is the combined mass of the frame and putty, $$x_f$$ is the initial stretch of the spring due to the weight of the frame, $$d$$ is the max distance due to collision the spring stretches, and $$v_0$$ is the initial velocity.

So now $$d$$ is the variable of a quadratic equation, and when I try to solve using the quadratic formula, I get imaginary numbers.

Edit: Attempted again with freshly made calculations. The results after plugging the numbers into the equation and attempting to solve with the quadratic formula resulted in real numbers, albeit negative.

 

[rwisz]

I would approach this problem by first considering the conservation of energy and momentum. In an inelastic collision, kinetic energy is not conserved, but momentum is. Therefore, we can use the conservation of momentum to solve for the final velocity of the putty and frame after the collision.

Let v be the final velocity of the combined mass of the frame and putty after the collision. We can set up the following equation based on the conservation of momentum:

mv = (m + M)v'

Where m is the mass of the putty, M is the mass of the frame, and v' is the final velocity of the combined mass after the collision. Solving for v', we get:

v' = mv / (m + M)

Next, we can use the fact that the spring is initially compressed by 0.050 m to determine the initial potential energy of the system. Using the equation for spring potential energy, we get:

U_o = (1/2)kx^2 = (1/2)(m + M)g(0.050)^2

Where k is the spring constant and x is the initial compression of the spring.

Now, we can use the conservation of energy to determine the final position of the combined mass after the collision. We know that the initial kinetic energy of the system is zero (since the putty is dropped from rest), and the final kinetic energy is (1/2)(m + M)v'^2. Therefore, we can set up the following equation:

U_o = (1/2)(m + M)v'^2 + (1/2)(m + M)gx^2

Solving for x, we get:

x = sqrt(2U_o / (m + M)g - v'^2)

Plugging in the values given in the problem, we get:

x = sqrt(2(0.015)(9.8)(0.050)^2 / (0.150 + 0.200) - (0.2(0.3)^2 / (0.150 + 0.200)^2)

x = 0.044 m

Therefore, the maximum distance the frame moves downward from its initial position is 0.044 m.