Question

Derive the relation between Final Velocity and Initial Velocity in the case of an Elastic Collision.

Chapter: Work, Power and Energy
Class 11, NCERT Text book

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Two bodies of mass M1 and M2 move in the same direction with velocities U1 and U2 collide (U1 > U2)after collision they velocity becomes V1 and V2.

$\huge{\bold{\underline{Explanation:-}}}$

As it is Elastic collision,

Both Momentum and Kinetic energy will be conserved.

Momentum (P) is conserved:-

$\large{\bold{P_i = P_f}}$

Now,

$\large{m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \: ----(1)}$

Kinetic energy (K.E) is conserved:-

$\large{\bold{K.E_i = K.E_f}}$

Now,

$\large{ \dfrac{1}{2}m_1(u_1)^2 + \dfrac{1}{2}m_2(u_2)^2 = \frac{1}{2}m_1(v_1)^2 + \dfrac{1}{2}m_2(v_2)^2}$

Half on each sides will get cancel

Then,

$\large{m_1(u_1)^2 + m_2(u_2)^2 = m_1(v_1)^2 + m_2(v_2)^2\: ----(2)}$

Now, From equation (1)

$\large{m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 }$

Rearranging it

$\large{m_1u_1 - m_1v_1 = m_2v_2 - m_2u_2 }$

$\large{m_1(u_1 - v_1) = m_2(v_2 - u_2) \: ----(3)}$

Now, From equation (2)

$\large{m_1(u_1)^2 + m_2(u_2)^2 = m_1(v_1)^2 + m_2(v_2)^2}$

Rearranging it

$\large{m_1(u_1)^2 - m_1(v_1)^2 = m_2(v_2)^2 - m_2(u_2)^2}$

$\large{m_1(u_1^2 - v_1^2) = m_2(v_2^2 - u_2^2) \: ----(4)}$

Now Divide equation (4) with equation (3)

Case-1

I.e

$\large{ \dfrac{equation \: 4}{equation \: 3}}$

$\large{ \dfrac{m_1(u_1^2 - v_1^2)}{m_1(u_1 - v_1)} = \dfrac{m_2(v_2^2 - u_2^2)}{m_2(v_2 - u_2)}}$

As we know,

$\large{a^2 - b^2 = (a+ b)(a-b)}$

$\large{ \dfrac{m_1[(u_1 - v_1) \times (u_1 + v_1)]}{m_1(u_1 - v_1)} = \dfrac{m_2[(v_2 - u_2) \times(v_2 + u_2)]}{m_2(v_2 - u_2)}}$

Now,

$\large{ \dfrac{m_1[ \cancel{(u_1 - v_1)} \times (u_1 + v_1)]}{m_1 \cancel{(u_1 - v_1)}} = \dfrac{m_2[ \cancel{(v_2 - u_2)} \times(v_2 + u_2)]}{m_2 \cancel{(v_2 - u_2)}}}$

It becomes,

$\large{u_1 + v_1 = v_2 + u_2}$

Now,

$\large{v_2 = v_1 + u_1 - u_2}$

Substituting the above equation In equation (1).

$\large{m_1u_1 + m_2u_2 = m_1v_1 + m_2(v_1 + u_1 - u_2)}$

$\large{m_1u_1 + m_2u_2 = m_1v_1 + m_2v_1 + m_2u_1 - m_2u_2}$

Rearranging,

$\large{m_1v_1 + m_2v_1 = m_1u_1 - m_2u_1 + m_2u_2 + m_2u_2}$

Now,

$\large{v_1(m_1 + m_2) = (m_1 - m_2)u_1 + 2m_2u_2}$

$\large{v_1 = \dfrac{(m_1 - m_2)u_1 + 2m_2u_2}{m_1 + m_2}}$

$\large{\boxed{\boxed{v_1 = \dfrac{(m_1 - m_2)u_1}{m_1 + m_2} + \dfrac{2m_2u_2}{m_1 + m_2}}}}$

Case-2

Now , take

$\large{v_1 = v_2 - u_1 + u_2}$

substituting the above equation in equation (1).

After simplifying,

we will get,

$\large{\boxed{\boxed{v_1 = \dfrac{(m_2 - m_1)u_1}{m_1 + m_2} + \dfrac{2m_1u_1}{m_1 + m_2}}}}$

Hence derived.

Note:-

In case-2 the procedure will be same that of case - 1.

Answer 2

$\huge{\star}{\underline{\boxed{\red{\sf{Answer :}}}}}{\star}$

We know that Momentum is Conserved.

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

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Kinetic Energy is also conserved

$\large{\therefore} \: {\sf{{\frac{1}{2}}m_{1}u_{1}^2 \: + \: {\frac{1}{2}}m_{2}u_{2}^2 \: = \: {\frac{1}{2}}m_{1}v_{1}^2 \: + \: {\frac{1}{2}}m_{2}v_{2}^2}}-------(2)$

===================>

From (1)

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

$\large{\implies}{\sf{m_{1}(u_{1} \: - \: v_{1}) \: = \: m_{2}(v_{2} \: - \: u_{1})}}------(3)$

=================>

From (2)

$\large{\sf{{\frac{1}{2}}m_{1}(u_{1}^2 \: - \: v_{1}^2) \: = \: {\frac{1}{2}}m_{2}(v_{2}^2 \: - \: u_{2}^2}})--------(4)$

>==============================<

Divide (4) by (3)

$\large{\sf{{\frac{u_{1}^2 \: - \: v_{1}^2}{u_{1} \: - \: v_{1}}} \: = \: {\frac{v_{2}^2 \: - \: u_{2}^2}{v_{2} \: - \: v_{1}}}}}$

$\large{\sf{u_{1} \: - \: u_{2} \: = \: v_{2} \: - \: v_{1}}}---------(5)$

====================>

From (5)

$\large{\sf{v_{1} \: = \: v_{2} \: - \: u_{1} \: + \: u_{2}}}---------(6)$

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As Momentum is conserved

$\large{\therefore}{\sf{m_{1}u_{1} \: + \: m_{2}u_{2} \: = \: m_{1}v_{1} \: + \: m_{2}v_{2}}}----------(7)$

>=============================<

Substite in (1)

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

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

$\large{\sf{2m_{1}u_{1} \: + \: (m_{2} \: - \: m_{1})u_{2} \: = \: (m_{1} \: + \: m_{2})v_{2}}}----(8)$

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For Final velocity :-

By using (8)

$\huge{\star}{\boxed{\blue{\sf{v_{2} \: = \: {\frac{2m_{1}u_{1} \: + \: (m_{2} \: - \: m_{1})u_{2}}{m_{1} \: + \: m_{2}}}}}}}$

$\rule{200}{2}$

Similarly for Initial Velocity :-

$\huge{\star}{\boxed{\blue{\sf{v_{1} \: = \: {\frac{2m_{2}u_{2} \: + \: (m_{1} \: - \: m_{2})u_{1}}{m_{2} \: + \: m_{1}}}}}}}$