Question
Prove that after a direct elastic collision between two balls of equal mass, their velocities mutually interchange.
✓
Answer
Suppose two balls of equal mass $m$ are moving in the same direction with velocities $u_1$ and $u_2$ respectively. Both the balls are moving in a same straight line before and after the collision and their collision is perfectly elastic. Let the velocities of the balls becomes $v_1$ and $v_2$ respectively after the collision.
Momentum and kinetic energy are conserved in elastic collisions, hence from the low of conservation of momentum,
Total momentum before collison $=$ Total momentum after collision
$\begin{aligned}
m u_1+m u_2 & =m v_1+m v_2 \\
u_1+u_2 & =v_1+v_2 \\
\left(u_1-v_1\right) & =\left(v_2-u_2\right) \ \ldots(1)
\end{aligned}$
From the law of conservation of kinetic energy, Kinetic energy before Collision $=$ kinetic energy after collision
$\begin{aligned}
\frac{1}{2} m u_1^2+\frac{1}{2} m u_2^2 & =\frac{1}{2} m v_1^2+\frac{1}{2} m v_2^2 \\
\left(u_1^2-v_1^2\right) & =\left(v_2^2-u_2^2\right) \ \ldots(2)
\end{aligned}$
By dividing equation (2) by (1),
$\begin{aligned}
\frac{\left(u_1^2-v_1^2\right)}{\left(u_1-v_1\right)} & =\frac{\left(v_2^2-u_2^2\right)}{\left(v_2-u_2\right)} \\
\frac{\left(u_1+v_1\right)\left(u_1-v_1\right)}{\left(u_1-v_1\right)} & =\frac{\left(v_2+u_2\right)\left(v_2-u_2\right)}{\left(v_2-u_2\right)} \\
u_1+v_1 & =v_2+u_2 \\
u_1-u_2 & =-\left(v_1-v_2\right) \ \ldots(3) \\
\Delta u & =-\Delta v
\end{aligned}$
Therefore, the relative speed of coming closer before the collision and the relative speed of moving away after the collision are equal.
On adding equation (1) and (3),
$\begin{aligned}
2 u_1 & =2 v_2 \\
\therefore u_1 & =v_2 \ \ldots(4)
\end{aligned}$
On subtracting equations (1) and (3)
$\begin{aligned}
2 u_2 & =2 v_1 \\
u_2 & =v_1 \ \ldots(5)
\end{aligned}$
Hence from equations (4) and (5) it is clear that after elastic collision between two moving balls of equal mass, their velocities mutually change.
Momentum and kinetic energy are conserved in elastic collisions, hence from the low of conservation of momentum,
Total momentum before collison $=$ Total momentum after collision
$\begin{aligned}
m u_1+m u_2 & =m v_1+m v_2 \\
u_1+u_2 & =v_1+v_2 \\
\left(u_1-v_1\right) & =\left(v_2-u_2\right) \ \ldots(1)
\end{aligned}$
From the law of conservation of kinetic energy, Kinetic energy before Collision $=$ kinetic energy after collision
$\begin{aligned}
\frac{1}{2} m u_1^2+\frac{1}{2} m u_2^2 & =\frac{1}{2} m v_1^2+\frac{1}{2} m v_2^2 \\
\left(u_1^2-v_1^2\right) & =\left(v_2^2-u_2^2\right) \ \ldots(2)
\end{aligned}$
By dividing equation (2) by (1),
$\begin{aligned}
\frac{\left(u_1^2-v_1^2\right)}{\left(u_1-v_1\right)} & =\frac{\left(v_2^2-u_2^2\right)}{\left(v_2-u_2\right)} \\
\frac{\left(u_1+v_1\right)\left(u_1-v_1\right)}{\left(u_1-v_1\right)} & =\frac{\left(v_2+u_2\right)\left(v_2-u_2\right)}{\left(v_2-u_2\right)} \\
u_1+v_1 & =v_2+u_2 \\
u_1-u_2 & =-\left(v_1-v_2\right) \ \ldots(3) \\
\Delta u & =-\Delta v
\end{aligned}$
Therefore, the relative speed of coming closer before the collision and the relative speed of moving away after the collision are equal.
On adding equation (1) and (3),
$\begin{aligned}
2 u_1 & =2 v_2 \\
\therefore u_1 & =v_2 \ \ldots(4)
\end{aligned}$
On subtracting equations (1) and (3)
$\begin{aligned}
2 u_2 & =2 v_1 \\
u_2 & =v_1 \ \ldots(5)
\end{aligned}$
Hence from equations (4) and (5) it is clear that after elastic collision between two moving balls of equal mass, their velocities mutually change.
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