How Do You Calculate the Initial Velocity of a Bullet in an Elastic Collision?

In summary, the problem involves a bullet of mass m colliding with a block of mass M on a rough surface with coefficient of kinetic friction k. The block slides for a distance of x meters before coming to a stop. The task is to derive an expression for the initial velocity of the bullet. To do this, we can use the principle of conservation of momentum, where the initial momentum of the bullet is equal to the initial momentum of the bullet plus block combination. This approach will allow us to calculate the initial velocity of the block and then use it to determine the initial velocity of the bullet.
  • #1
spectravoid
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Homework Statement


A bullet of mass m is shot into a block of mass M that rests on a rough surface with coefficient of kinetic friction k. The block slides for a distance of x meters before coming to a stop. Derive an expression for the initial velocity of the bullet.

Homework Equations


0.5mvi^2 - k(-1)x = 0.5(M+m)vf^2 (work energy theorem just with the variables replaced with the ones in the question)
and also this but I'm not sure where to use it Ff = (M + m)gk

The Attempt at a Solution


vi = +/- sqrt((vf^2 + (Mvf^2)/m + (2kx)/m))

I need to remove vf

thank you for help
 
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  • #2
You don't want to use the work/energy stuff here. This is an INelastic collision. When the bullet collides with the block it is stopped by friction within the block and you have no data to calculate that directly. What you can count on is conservation of momentum. The initial momentum of the bullet is equal to the initial momentum of the bullet plus block combination. This will let you get an initial velocity for the block. Go from there.
 
  • #3


I would approach this problem by first reviewing the given information and equations. The work-energy theorem states that the initial kinetic energy of the bullet (0.5mvi^2) must be equal to the final kinetic energy of the bullet and block (0.5(M+m)vf^2) minus the work done by friction (k(-1)x). We can also use the equation Ff = (M + m)gk to calculate the force of friction acting on the block.

To derive an expression for the initial velocity of the bullet, we can rearrange the work-energy theorem equation to solve for vi:

0.5mvi^2 = 0.5(M+m)vf^2 - k(-1)x

We can then substitute the equation for force of friction into the work-energy theorem:

0.5mvi^2 = 0.5(M+m)vf^2 - (M + m)gk(-1)x

Next, we can substitute the equation for final velocity (vf) from the first equation into the second equation:

0.5mvi^2 = 0.5(M+m)((0.5mvi^2 + (Mvi^2)/m + (2kx)/m)) - (M + m)gk(-1)x

Simplifying and solving for vi, we get the following expression for the initial velocity of the bullet:

vi = +/- sqrt((vf^2 + (Mvf^2)/m + (2kx)/m))

This expression gives us the initial velocity of the bullet based on the final velocity, mass of the block and bullet, coefficient of kinetic friction, and distance traveled by the block. We can use this equation to calculate the initial velocity for different scenarios and analyze the effects of changing variables such as mass, friction, and distance.
 

1. What is an elastic collision?

An elastic collision is a type of collision between two objects where no energy is lost. This means that the total kinetic energy of the system before and after the collision remains the same.

2. How is the momentum conserved in an elastic collision?

In an elastic collision, the total momentum of the system is conserved. This means that the sum of the momentum of the two objects before the collision is equal to the sum of their momentum after the collision.

3. What are the equations used to solve elastic collision problems?

The equations used to solve elastic collision problems are the conservation of momentum equation and the conservation of kinetic energy equation. These equations can be used to solve for the final velocities of the objects involved in the collision.

4. Can an elastic collision occur between two objects of different masses?

Yes, an elastic collision can occur between two objects of different masses. The final velocities of the objects will depend on their masses and initial velocities, but the total kinetic energy of the system will remain the same.

5. How does an elastic collision differ from an inelastic collision?

An elastic collision differs from an inelastic collision in that in an inelastic collision, some of the kinetic energy is lost and is converted into other forms of energy, such as heat. In an elastic collision, no energy is lost and the total kinetic energy remains the same before and after the collision.

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