# What is the maximum spring compression in a bullet and block collision?

• huskydc
In summary, a projectile with a mass of 20 g and an initial horizontal velocity of 100 m/s hits and stops in a wood block with a mass of 0.403 kg. The block is attached to a spring with a spring constant of 148 N/m. By using conservation of momentum, the final velocity of the block is determined to be 4.963 m/s. To find the maximum compression of the spring, the relationship between energy and velocity is used, but the mass of the bullet must also be considered.
huskydc
A projectile of mass 20 g has an initial horizontal velocity of 100 m/s when it hits and stops in a wood block of mass 0.403 kg. The block is sitting on a horizontal frictionless surface and is attached to a massless spring, initially relaxed, with spring constant 148 N/m. What is the maximum compression of the spring?

since momentum is conserved, i used the equation
mv(initial) = mv (final), and found the final velocity of the block, as a result of the mass hitting it, is 4.963,

but I'm not sure where to go from here? ..or am i heading the right direction ?

You have the velocity of the block. Now you can use energy to solve for the compression of the spring. (KE = .5mv^2; Es=.5kx^2)
You're on the right track. Just try to think of the relationships between the various quantities (often there is more than one, i.e. velocity is related to momentum, but it is also related to energy)

ok, I've set up the equations as follow:

.5kx^2 = KE (in this case, also KE final)
thus,

.5kx^2 = .5m(4.963^2)

and solve for x... but still incorrect, something wrong with my equations? or are there other factors i haven't thought of?

huskydc said:
ok, I've set up the equations as follow:

.5kx^2 = KE (in this case, also KE final)
thus,

.5kx^2 = .5m(4.963^2)

and solve for x... but still incorrect, something wrong with my equations? or are there other factors i haven't thought of?
Make sure you include the mass of the bullet since it embeds in the block.

## 1. What is the purpose of a bullet, block, and spring?

The purpose of a bullet, block, and spring is to demonstrate the principles of conservation of momentum and energy in physics. It is a common physics demonstration used in classrooms to show how energy is transferred between objects during collisions.

## 2. How does a bullet, block, and spring demonstration work?

In this demonstration, a bullet is fired into a stationary block attached to a spring. The bullet's momentum is transferred to the block, causing it to move and compress the spring. The energy stored in the spring is then released, causing the block to bounce back and the bullet to be launched in the opposite direction.

## 3. What factors affect the outcome of a bullet, block, and spring demonstration?

The outcome of this demonstration can be affected by several factors, including the mass and velocity of the bullet, the stiffness of the spring, and the friction between the block and surface it is resting on. These factors can impact the amount of energy transferred and the final velocities of the bullet and block.

## 4. What are the real-life applications of the bullet, block, and spring principle?

The principles demonstrated in this experiment are applicable in many real-life situations, such as car accidents, sports collisions, and even in the design of airbags. Understanding how energy is transferred during collisions is crucial in creating safety measures to protect individuals from harm.

## 5. Can the bullet, block, and spring demonstration be used to calculate values?

Yes, this demonstration can be used to calculate values such as the bullet's initial velocity, the spring's stiffness, and the amount of energy transferred. However, it is important to note that these calculations are idealized and may not accurately reflect real-world scenarios due to factors such as air resistance and imperfections in the materials used.

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