# Hooke's Law Problem: Finding Maximum Displacement with Inelastic Collision

• sr57
In summary, a 0.2 kg block of wood attached to a spring with a spring constant of 25 N/m undergoes an inelastic collision with a 0.05 kg dart traveling at 0.1 m/s parallel to the x-axis. The final velocity of the block is 0.02 m/s. To find the maximum displacement of the block from its equilibrium position, the sum of elastic potential energy and kinetic energy must be conserved. Using this, the maximum displacement is found to be 0.01 m.

#### sr57

A 0.2 kg block of wood is attached to a spring with a spring constant k = 25 N/m. The block is initially at rest and the spring is at its equilibrium length aligned along the x-axis. A dart of mass 0.05 kg is thrown at a block of wood, undergoes an inelastic collision and sticks into the block. the initial speed of the dart is 0.1 m/s and is parallel to the x-axis. The maximum displacement of the block from its equilibrium position in m is: ?

## The Attempt at a Solution

I used m1Vi + m2Vi = (m1+ m2) Vf
to find Vf = 0.1 m/s

I know F= k x
I don't know know how you would find F.

Please recheck your vf, because I find something different using your equation. Then, you can try using Hooke's law with conservation of energy (the spring force is a conservative force). Where is the energy from the moving block and dart going when they slow down to a stop?

I don't think you have the right value for v_f -

(0.2 kg)*(0 m/s) + (0.05 kg)*(0.1 m/s) = (0.25 kg)*v_f
v_f = 0.02 m/s

Although the collision between the dart and the block is inelastic, the compression of a spring does conserve the sum of the kinetic energy and the elastic potential energy!

Thank you I got it. I made a mistake with the calculations b4..but this is the way right?

Since F= PV => F = (mv)v ==> F= 0.25

F= kx
0.25/25 N/m = x
x= 0.01 m

Why do you say F = pv?

F is not equal to the product of momentum and velocity... (you can see this by comparing the units of F (kg*m/s^2) and the units of pv (kg*m/s*m/s)).

It's important to see here that the sum of elastic potential energy and kinetic energy is conserved. So what is the elastic potential energy of a spring compressed distance x?

Sorry I meant P = Fv therefore F = P/V and since P=mv, F = mv/v = m ?

Whoa, you've just said F=mv/v = m... But force isn't equal to mass!

oedipa maas said:
F is not equal to the product of momentum and velocity... (you can see this by comparing the units of F (kg*m/s^2) and the units of pv (kg*m/s*m/s)).

It's important to see here that the sum of elastic potential energy and kinetic energy is conserved. So what is the elastic potential energy of a spring compressed distance x?

So 1/2 mv^2 = 1/2 kx^2
x = 0.01 m

Can you show your substitution? There seems to be a mistake there.

1/2 (0.25 kg)(0.02 m/s) = 1/2 (25 N/m)(x^2)
x^2 = 0.002 m

I put the answer for the rong question before..sorry abt that :\$

## 1. What is Hooke's Law?

Hooke's Law is a principle in physics that states the amount of force needed to extend or compress a spring is directly proportional to the distance it is stretched or compressed. This law is often represented as F = kx, where F is the force applied, k is the spring constant, and x is the displacement of the spring.

## 2. How is Hooke's Law related to an inelastic collision?

In an inelastic collision, kinetic energy is not conserved as some of the energy is converted to other forms such as heat or sound. Hooke's Law can be applied to determine the maximum displacement of a spring involved in an inelastic collision, as it relates the force applied to the displacement of the spring.

## 3. How do you calculate the maximum displacement using Hooke's Law?

To calculate the maximum displacement using Hooke's Law, you will need to know the mass and velocity of the object before and after the collision, as well as the spring constant of the spring involved. Using the equation F = kx, you can rearrange to solve for x, which will give you the maximum displacement.

## 4. Can Hooke's Law be used for other types of collisions?

Hooke's Law can only be applied to collisions where a spring is involved. For other types of collisions, different principles and laws must be used to determine variables such as displacement, velocity, and force.

## 5. What are some real-life applications of Hooke's Law?

Hooke's Law has many practical applications in fields such as engineering, construction, and sports. Some examples include the design of suspension systems in cars, the construction of buildings and bridges to withstand external forces, and the calculation of the optimal spring stiffness for sports equipment such as tennis rackets and diving boards.