Question

consider a system of two masses m1 and m2. let u1 be the initial velocity of mass m1 and let m2 be at rest. the masses undergo an elastic collision in one dimension. derive the expression for the velocities of the two masses after the collision.​

Answer 1

Given : Consider a system of two masses m₁ and m₂

let u₁ be the initial velocity of mass m₁ and let m₂ be at rest.

the masses undergo an elastic collision in one dimension.

To Find : derive the expression for the velocities of the two masses after the collision.​

Solution:

mass  m₁  , initial velocity u₁

mass  m₂  , initial velocity 0  ( at rest )

Momentum = m₁u₁ + m₂(0)   = m₁u₁

KE = (1/2)m₁u₁² + (1/2)m₂(0)² = (1/2)m₁u₁²

Let say

mass  m₁  , Final velocity v₁

mass  m₂  , Final velocity v₂  

Momentum = m₁v₁ + m₂v₂

KE = (1/2)m₁v₁² + (1/2)m₂(v₂) ²  

Using conservation of momentum

m₁v₁ + m₂v₂ = m₁u₁

=> v₂ = m₁(u₁ - v₁)/m₂

As collision is elastic hence KE is conserved

(1/2)m₁v₁² + (1/2)m₂(v₂) ²   = (1/2)m₁u₁²

=> m₁v₁² + m₂(v₂) ²   = m₁u₁²

Substituting v₂ = m₁(u₁ - v₁)/m₂

m₁v₁² + m₂(m₁(u₁ - v₁)/m₂) ²   =  m₁u₁²

=> m₁m₂v₁² +  m₁²(u₁ - v₁) ²   =  m₁m₂u₁²

=> m₂v₁² +  m₁(u₁ - v₁)(u₁ - v₁)    =   m₂u₁²

=>   m₁(u₁ - v₁)(u₁ - v₁)  =  m₂(u₁² - v₁² )

=>   m₁(u₁ - v₁)(u₁ - v₁)  =  m₂(u₁  + v₁  )(u₁  - v₁  )

=>   m₁(u₁ - v₁)  =  m₂(u₁  + v₁  )

=>  m₁ u₁ -  m₁v₁   =  m₂ u₁  + m₂ v₁

=> u₁ (m₁ -m₂) = v₁ (m₁ +m₂)

=>  v₁  = u₁ (m₁ -m₂)/(m₁ +m₂)

v₂ = m₁(u₁ - v₁)/m₂

=> v₂ = m₁(u₁ - u₁ (m₁ -m₂)/(m₁ +m₂))/m₂

=> v₂ = m₁( 2m₂u₁)/m₂(m₁ +m₂)

=> v₂ = 2m₁u₁/(m₁ +m₂)

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Answer 2

Given:

Consider a system of two masses m1 and m2. Let u1 be the initial velocity of mass m1 and let m2 be at rest. The masses undergo an elastic collision in one dimension.

To find:

Final velocities of each masses ?

Calculation:

We will use two tricks to solve this question easily !

• Conservation of Linear Momentum
• Coefficient of Restitution

Conservation of Linear Momentum:

$\sf \therefore \: m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$

$\sf \implies\: m_{1}u_{1} + m_{2}(0) = m_{1}v_{1} + m_{2}v_{2}$

$\sf \implies\: m_{1}u_{1} = m_{1}v_{1} + m_{2}v_{2}$

According to Coefficient of Restitution:

$\sf \therefore \: e = \dfrac{v_{2} -v_{1}}{ - (0 -u_{1})}$

$\sf \implies \: e = \dfrac{v_{2} -v_{1}}{ u_{1}}$

$\sf \implies \: 1 = \dfrac{v_{2} -v_{1}}{ u_{1}} \: \: \: \: . \: . \: .(e = 1 \: for \: elastic \: collision)$

$\sf \implies \: v_{2} -v_{1} = u_{1}$

__________________________________

Now, substituting value of v2 in 1st eq:

$\sf \implies\: m_{1}u_{1} = m_{1}v_{1} + m_{2}v_{2}$

$\sf \implies\: m_{1}u_{1}= m_{1}v_{1} + m_{2}(u_{1} + v_{1})$

$\sf \implies\: (m_{1} - m_{2})u_{1} = (m_{1} + m_{2})v_{1}$

$\sf \implies\:v_{1} = \dfrac{(m_{1} - m_{2})}{(m_{1} + m_{2})}u_{1}$

_______________________

Now, substituting v1 in 1st eq:

$\sf \implies\: m_{1}u_{1}= m_{1}v_{1} + m_{2}v_{2}$

$\sf \implies\: m_{1}u_{1} + m_{2}u_{2} = m_{1}(v_{2} - u_{1} )+ m_{2}v_{2}$

$\sf \implies\: m_{1}u_{1}= m_{1}(v_{2} - u_{1} )+ m_{2}v_{2}$

$\sf \implies\: 2m_{1}u_{1}= m_{1}v_{2} + m_{2}v_{2}$

$\sf \implies\: 2m_{1}u_{1}= (m_{1}+ m_{2})v_{2}$

$\sf \implies\: v_{2} = \dfrac{2m_{1}u_{1}}{(m_{1} +m_{2}) }$

Hope It Helps.