# Particles after an elastic collision

• Suyash Singh
In summary, the Homework Equations state that: u initial velocity v final velocity The Attempt at a Solution states that: after collision A->V-x B->x λA/λB=(h/m(v-x)) / (h/( m/2 x)) = 1/2x/v-x momentum conservation-- mv=m(v-x)+ m/2(x) v=v-x+x/2 x=0 ! now what??I really don't understand anything and all sites do different solutions so please don
Suyash Singh

## Homework Statement

A particle A of mass m and initial velocity v collides with a particle B of mass m 2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is

## Homework Equations

u initial velocity
v final velocity

## The Attempt at a Solution

after collision
A->V-x
B->x
λA/λB=(h/m(v-x)) / (h/( m/2 x)) = 1/2x/v-x

momentum conservation--
mv=m(v-x)+ m/2(x)
v=v-x+x/2

x=0 !

now what??
I really don't understand anything and all sites do different solutions so please don't put my question in black hole.

Last edited by a moderator:
Suyash Singh said:

## Homework Statement

A particle A of mass m and initial velocity v collides with a particle B of mass m 2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is

## Homework Equations

u initial velocity
v final velocity

## The Attempt at a Solution

after collision
A->V-x
B->x
λA/λB=(h/m(v-x)) / (h/( m/2 x)) = 1/2x/v-x

momentum conservation--
mv=m(v-x)+ m/2(x)
v=v-x+x/2

x=0 !

now what??
I really don't understand anything and all sites do different solutions so please don't put my question in black hole.

you are doing elastic collision
so the momentum and energy both will be conserved.
write down energy before the collision and after the collision and equate them
similarly for momentum ...before and after will be equal

solve for velocities of the two particles after collision. and then write down the wavelength = h/momentum and take ratio.

Suyash Singh
For perfectly elastic collisions you can use the formula Δp = 2μΔv where μ is the reduced mass [ m1 * m2 /(m1 + m2) ] of the colliding objects and Δv is their relative velocity. Then momentum of stationary object after collision will be Δp and of moving object p(before) - Δp. Convert to wavelengths using the normal formula.

## 1. What is an elastic collision?

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

## 2. What happens to the particles after an elastic collision?

The particles involved in an elastic collision will continue moving after the collision, but their velocities and directions may change. In an elastic collision, the total momentum of the particles is conserved, meaning that the sum of the initial momenta of the particles is equal to the sum of the final momenta.

## 3. How do you calculate the velocities of particles after an elastic collision?

The velocities of particles after an elastic collision can be calculated using the conservation of momentum and conservation of kinetic energy equations. These equations take into account the masses and initial velocities of the particles, as well as the angle of collision.

## 4. Are there any real-life examples of elastic collisions?

Yes, there are many real-life examples of elastic collisions. One common example is billiard balls colliding on a pool table. The collision between the balls is considered elastic because no energy is lost during the collision.

## 5. What is the difference between an elastic collision and an inelastic collision?

In an elastic collision, no energy is lost and the total kinetic energy of the particles remains the same. In an inelastic collision, some energy is lost, usually in the form of heat or sound, and the total kinetic energy of the particles decreases. Inelastic collisions are often described as "sticky" collisions, while elastic collisions are described as "bouncy" collisions.

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