A Particle Of Mass m1 Moving With Velocity u1 Collides Head-On Collision With A Particle Of Mass m2 Moving With Velocity u2. If ‘e’ Be The Coefficient Of Restitution, Then Calculate The Loss Of The Kinetic Energy As A Result Of The Collision.

Ans.

A particle of mass \( m_1 \) moving with velocity \( u_1 \) collides head-on collision with the particle of mass \( m_2 \) moving with the velocity \( u_2 \).

Let us consider \( v_1 \) and \( v_2 \) be the velocities of mass \( m_1 \) and \( m_2 \) respectively.

According to the conservation of linear momentum,

\( m_1u_1+m_2u_2=m_1v_1+m_2v_2 \tag{1}\)

If \( e \) be the coefficient of restitution, then

\( \displaystyle{e=\frac{v_2-v_1}{u_1-u_2}}\tag{2} \)

The loss of kinetic energy due to the impact is

=Kinetic energy before collision – Kinetic energy after collision

= \( (\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2)-(\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2) \)

= \( \frac{1}{2}\left[(m_1u_1^2+m_2u_2^2)-(m_1v_1^2+m_2v_2^2)\right] \)

= \( \frac{1}{2(m_1+m_2)}\left[(m_1+m_2)(m_1u_1^2+m_2u_2^2)-(m_1+m_2)(m_1v_1^2+m_2v_2^2)\right] \)

= \( \frac{1}{2(m_1+m_2)}\left[\left(m_1^2u_1^2+m_2^2u_2^2+m_1m_2(u_1^2+u_2^2)\right)\\-\left(m_1^2v_1^2+m_2^2v_2^2+m_1m_2(v_1^2+v_2^2)\right) \right] \)

= \( \frac{1}{2(m_1+m_2)}\left[\left({(m_1u_1+m_2u_2)}^2-2m_1m_2u_1u_2+m_1m_2(u_1^2+u_2^2)\right)\\-\left({(m_1v_1+m_2v_2)}^2-2m_1m_2v_1v_2+m_1m_2(v_1^2+v_2^2)\right) \right] \)

= \( \frac{1}{2(m_1+m_2)}\left[\left({(m_1u_1+m_2u_2)}^2+m_1m_2(u_1^2+u_2^2-2u_1u_2)\right)\\-\left({(m_1v_1+m_2v_2)}^2+m_1m_2(v_1^2+v_2^2-2v_1v_2)\right) \right] \)

= \( \frac{1}{2(m_1+m_2)}\left[{(m_1u_1+m_2u_2)}^2+m_1m_2{(u_1-u_2)}^2\\-{(m_1v_1+m_2v_2)}^2-m_1m_2{(v_1-v_2)}^2\right] \)

= \( \frac{1}{2(m_1+m_2)}\left[m_1m_2{(u_1-u_2)}^2-m_1m_2{(v_1-v_2)}^2\right] \)
[ using equation (1)]

= \( \frac{1}{2(m_1+m_2)}\left[m_1m_2{(u_1-u_2)}^2m_1m_2e^2{(u_1-u_2)}^2\right] \)
[using equation (2)]

= \( \displaystyle{\frac{m_1m_2}{2(m_1+m_2)}{(u_1-u_2)}^2(1-e^2)} \)

CASE-I:
In case of a perfectly inelastic collision \( e=0 \),
So the loss of the kinetic energy will be
\( \displaystyle{\frac{m_1m_2}{2(m_1+m_2)}{(u_1-u_2)}^2} \)

CASE-II:
In case of perfectly elastic collision \( e=1 \),
So the loss of the kinetic energy will be
\( 0 \)
This means that after collision the total kinetic energy will remain unchanged.

Ans.

You Might Also Like

Prove That The Total Momentum Of A System Is Constant i.e., Conserved Then The Centre Of Mass Is Either At Rest Or In Motion With Constant Velocity.

Show That The Sum Of The Products Of Mass And Position Vectors Of All Particles i.e., The Sum Of Moments Of Mass Points About The Centre Of Mass Is Zero.

A Rectangular Plate Having Edges Of Lengths ‘a’ And ‘b’ Respectively, Hangs Vertically From The Edge Of Length ‘a’. (i) Find The Time Period For Small Oscillations And (ii) The Length Of The Equivalent Simple Pendulum.