# How to find resulting velocity in a perfectly elastic collision?

• haha0p1
In summary, the principle of conservation of momentum states that the total momentum of two particles after a collision is equal to the sum of the initial and final momentum of each particle.
haha0p1
Homework Statement
A particle of mass m travelling with velocity u collides elastically and head-on with a stationary particle of mass M. Which expression gives the velocity of the particle of mass M after the collision.
Relevant Equations
Momentum=Mass×Velocity
Using principle of conservation of momentum:
m×u=m×v1 + M×v2
Where m=mass of moving particle in the beginning
u=Initial velocity of particle m
v1= final velocity of particle m
v2=velocity of object M
m×u-(mv1)=Mv2
(mu-mv1)÷M=v2
My answer is this (mu-mv1)÷M
However, it is nowhere close to the correct answer. Kindly tell where I am going wrong in the calculation.

You should not have v1 in the answer since that is also unknown. You have not used that the collision is perfectly elastic.

haruspex said:
You should not have v1 in the answer since that is also unknown. You have not used that the collision is perfectly elastic.
I have used V1 im the equation because the initial momentum is equal to final momentum and V1 is a part of the final momentum.

haha0p1 said:
I have used V1 im the equation because the initial momentum is equal to final momentum and V1 is a part of the final momentum.
There's nothing wrong with it as an equation, but it is not acceptable as an answer. Only m, M and u are allowed. To eliminate v1 you need another equation, the equation for energy conservation.

MatinSAR
Using principle of conservation of momentum:
mu=(m×v-u)+M×v
mu-m(v-u)=M×v
-mv÷M=v

Note:
m=mass of the initially moving object
v=velocity of object woth mass M
x-u=Velocity of object m after the collision

I have used a different method but I am still getting a wrong answer.

If the different method does not include kinetic energy conservation as @haruspex suggested, you will keep getting a wrong answer. The initial momentum conservation equation you had
haha0p1 said:
m×u-(mv1)=Mv2
is correct so leave it alone. Be sure to use subscripts to avoid confusion.

MatinSAR

## 1. What is a perfectly elastic collision?

A perfectly elastic collision is a type of collision in which there is no loss of kinetic energy. This means that the total kinetic energy of the system before and after the collision remains the same.

## 2. How do you calculate the resulting velocity in a perfectly elastic collision?

The resulting velocity in a perfectly elastic collision can be calculated using the conservation of momentum and the conservation of kinetic energy equations. The formula for the resulting velocity is:
vf = (m1 * v1i + m2 * v2i) / (m1 + m2)
where vf is the resulting velocity, m1 and m2 are the masses of the two objects involved in the collision, and v1i and v2i are the initial velocities of the two objects.

## 3. Can the resulting velocity in a perfectly elastic collision be negative?

Yes, the resulting velocity in a perfectly elastic collision can be negative if one of the objects involved in the collision is moving in the opposite direction of the other object. This negative velocity indicates that the object is moving in the opposite direction after the collision.

## 4. What is the difference between a perfectly elastic collision and an inelastic collision?

A perfectly elastic collision is a type of collision in which there is no loss of kinetic energy, while an inelastic collision is a type of collision in which there is a loss of kinetic energy. In an inelastic collision, the total kinetic energy of the system before the collision is greater than the total kinetic energy after the collision.

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

Yes, there are real-life examples of perfectly elastic collisions. One example is the collision between two billiard balls on a pool table. Another example is the collision between two atoms or molecules in a gas, where there is no loss of kinetic energy.

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