Question

Two spheres are placed in a horizontal plane with kinetic energies $(KE{)}_A=8\text{J}$ and $(KE{)}_B=18\text{J}$ as shown in figure. If both the spheres collide elastically, find the speed of both the spheres after collision. Both the spheres have the same mass $m=1\text{kg}$.

• A. $V_A=6\text{m/s}{V}_B=4\text{m/s}$
• B. $V_A=4\text{m/s}{V}_B=6\text{m/s}$
• C. $V_A=3\text{m/s}{V}_B=2\text{m/s}$
• D. $V_A=2\text{m/s}{V}_B=3\text{m/s}$

Answer 1

The correct option is A $V_A=6\text{m/s}{V}_B=4\text{m/s}$

As we know, in elastic collision, kinetic energy of the system remains same before and after the collision.
Hence, after collision, total K.E $=26\text{J}$
Initial speeds before collision:
For $\left(B\right)$, $\frac{1}{2}m{u}_2^2=18\Rightarrow u_2=6\text{m/s}$
For $\left(A\right)$, $\frac{1}{2}m{u}_1^2=8\Rightarrow u_1=4\text{m/s}$


Let the speeds of $A$ and $B$ after collision be $v_1$ and $v_2$.
Then, kinetic energy after collision
$K.E=\frac{1}{2}m{v}_2^2+\frac{1}{2}m{v}_1^2=26$
or $v_1^2+v_2^2=52...\left(i\right)$

Collision is elastic hence, $e=1$
$1=e=\frac{v_1-v_2}{6-4}\Rightarrow v_1-v_2=2\text{m/s}...\left(ii\right)$
From equation (i) and (ii)
$v_2=4\text{m/s}$
$v_1=6\text{m/s}$