Question

please help 8.) Two identical balls, A and B collide head on elastically. If
the velocity of A and B before collision are 0.5 ms and
0.3 ms respectively, then their velocities after collision
will be
Kerala CEEI
(a) 0.5 ms - and 0.3 ms -
(b) - 0.5 ms and 0.3 ms -
(c) 0.3 ms -and - 0.5 ms
(d) 0.3 ms - and 0.5 ms -
(e) - 0.3 ms - and 0.5 ms -1​

Answer 1

Mass of balls A and B are same.

$\displaystyle\longrightarrow\sf{m_A=m_B=m\ (Say)}$

Initial velocity of ball A, $\displaystyle\sf {u_A=0.5\ m\,s^{-1}}$

Initial velocity of ball B, $\displaystyle\sf {u_B=-0.3\ m\,s^{-1}}$

Negative sign is because both are moving towards each other so one moves opposite to the other, since head - on collision takes place.

Let $\displaystyle\sf {v_A}$ and $\displaystyle\sf {v_B}$ be final velocity of balls A and B respectively.

Ball A moves towards its initial position after collision, i.e., it reverses it's motion. So does B.

Thus, since $\displaystyle\sf {u_A\ \textgreater\ 0}$ and $\displaystyle\sf {u_B\ \textless\ 0,}$

• $\displaystyle\sf {v_A\ \textless\ 0}$

• $\displaystyle\sf {v_B\ \textgreater\ 0}$

By conservation of linear momentum,

$\displaystyle\longrightarrow\sf{m_Au_A+m_Bu_B=m_Av_A+m_Bv_B}$

$\displaystyle\longrightarrow\sf{0.5m-0.3m=mv_A+mv_B}$

$\displaystyle\longrightarrow\sf{0.2m=m(v_A+v_B)}$

$\displaystyle\longrightarrow\sf{v_A+v_B=0.2\quad\quad\dots (1)}$

Since the collision is elastic, kinetic energy is conserved.

$\displaystyle\longrightarrow\sf{\dfrac {1}{2}\,m_A(u_A)^2+\dfrac {1}{2}\,m_B(u_B)^2=\dfrac {1}{2}\,m_A(v_A)^2+\dfrac {1}{2}\,m_B(v_B)^2}$

$\displaystyle\longrightarrow\sf{\dfrac {1}{2}\,m(0.5)^2+\dfrac {1}{2}\,m(-0.3)^2=\dfrac {1}{2}\,m(v_A)^2+\dfrac {1}{2}\,m(v_B)^2}$

$\displaystyle\longrightarrow\sf{\dfrac {1}{2}\,m\left [0.25+0.09\right]=\dfrac {1}{2}\,m\left [(v_A)^2+(v_B)^2\right]}$

$\displaystyle\longrightarrow\sf{(v_A)^2+(v_B)^2=0.34\quad\quad\dots (2)}$

Squaring (1),

$\displaystyle\longrightarrow\sf{(v_A+v_B)^2=(0.2)^2}$

$\displaystyle\longrightarrow\sf{(v_A)^2+2v_Av_B+(v_B)^2=0.04\quad\quad\dots (3)}$

From (2),

$\displaystyle\longrightarrow\sf{0.34+2v_Av_B=0.04}$

$\displaystyle\longrightarrow\sf{2v_Av_B=-0.30}$

$\displaystyle\longrightarrow\sf{4v_Av_B=-0.60\quad\quad\dots (4)}$

Subtracting (4) from (3),

$\displaystyle\longrightarrow\sf{(v_A)^2+2v_Av_B+(v_B)^2-4v_Av_B=0.04+0.60}$

$\displaystyle\longrightarrow\sf{(v_A)^2-2v_Av_B+(v_B)^2=0.64}$

$\displaystyle\longrightarrow\sf{(v_A-v_B)^2=0.64}$

We see $\displaystyle\sf {v_A-v_B\ \textless\ 0.}$ Thus,

$\displaystyle\longrightarrow\sf{v_A-v_B=-0.8\quad\quad\dots (5)}$

On adding (1) and (5),

$\displaystyle\longrightarrow\sf{\underline {\underline {v_A=-0.3\ m\,s^{-1}}}}$

On subtracting (5) from (1),

$\displaystyle\longrightarrow\sf{\underline {\underline {v_B=0.5\ m\,s^{-1}}}}$

Hence (e) is the answer.

But this answer is possible if we consider motion of ball A positive and that of ball B negative.

If motion of ball A is taken negative and that of B positive, then (c) will be the answer, because there occurs change in sign of velocities too.