How to find resulting velocity in a perfectly elastic collision?

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.
  • #1
haha0p1
46
9
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.
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  • #2
You should not have v1 in the answer since that is also unknown. You have not used that the collision is perfectly elastic.
 
  • #3
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.
 
  • #4
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.
 
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  • #5
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.
 
  • #6
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.
 
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FAQ: How to find resulting velocity in a perfectly elastic collision?

1. What is a perfectly elastic collision?

A perfectly elastic collision is a type of collision where there is no loss of kinetic energy in the system. Both momentum and kinetic energy are conserved during the collision.

2. How do you find the resulting velocity of two objects after a perfectly elastic collision?

To find the resulting velocities of two objects after a perfectly elastic collision, you can use the following formulas:\[ v_{1f} = \frac{(m_1 - m_2) v_{1i} + 2m_2 v_{2i}}{m_1 + m_2} \]\[ v_{2f} = \frac{(m_2 - m_1) v_{2i} + 2m_1 v_{1i}}{m_1 + m_2} \]where \( v_{1i} \) and \( v_{2i} \) are the initial velocities of objects 1 and 2, \( v_{1f} \) and \( v_{2f} \) are the final velocities, and \( m_1 \) and \( m_2 \) are the masses of the objects.

3. What are the conservation laws used in solving perfectly elastic collisions?

In a perfectly elastic collision, both the conservation of momentum and the conservation of kinetic energy are used. The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. The conservation of kinetic energy states that the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

4. Can you provide an example problem and solution for finding the resulting velocities in a perfectly elastic collision?

Sure! Suppose we have two objects: Object 1 with mass \( m_1 = 2 \) kg and initial velocity \( v_{1i} = 3 \) m/s, and Object 2 with mass \( m_2 = 3 \) kg and initial velocity \( v_{2i} = -2 \) m/s. Using the formulas:\[ v_{1f} = \frac{(2 - 3) \cdot 3 + 2 \cdot 3 \cdot (-2)}{2 + 3} = \frac{-3 + (-12)}{5} = \frac{-15}{5} = -3 \text{ m/s} \]\[ v_{2f} = \frac{(3 - 2) \cdot (-2) + 2 \cdot 2 \cdot 3}{2 + 3} = \frac{(-2) + 12}{5} = \frac{10}{5} = 2 \text

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