Question

In an elastic collision between spheres $A$ and $B$ of equal mass but unequal radii, $A$ moves along the $x-$ axis and $B$ is stationary before impact. Which of the following is possible after impact?

• A. $A$ comes to rest.
• B. The velocity of $B$ relative to $A$ remains the same in magnitude but reverses in direction.
• C. $A$ and $B$ move with equal speeds, making an angle of ${45}^o$ each with the $x-$ axis.
• D. $A$ and $B$ move with unequal speeds, making angles of ${30}^o$ and ${60}^o$ with $x-$ axis respectively.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $A$ comes to rest.
B The velocity of $B$ relative to $A$ remains the same in magnitude but reverses in direction.
C $A$ and $B$ move with equal speeds, making an angle of ${45}^o$ each with the $x-$ axis.
D $A$ and $B$ move with unequal speeds, making angles of ${30}^o$ and ${60}^o$ with $x-$ axis respectively.

Option (A)
A comes to rest and B moves with the velocity of B.
Option(B):
A comes to rest and B moves with the velocity of B.
Therefore, velocity of B relative to A remains the same in magnitude but reverses in direction which follows from momentum conservation.
Option (C):
$mv+0=m{v}^{\prime }\cos {45}^0+m{v}^{\prime }\cos {45}^0=2m{v}^{\prime }\cos {45}^0=\sqrt{2}m{v}^{\prime }$
$\Rightarrow v=\sqrt{2}{v}^{\prime }$ This is possible.
Option (D):
$mv=m{v}_1\cos {30}^0+m{v}_2\cos {60}^0=m{v}_1\times \frac{\sqrt{3}}{2}+m{v}_2\frac{1}{2}$
$\Rightarrow 2v=\sqrt{3}{v}_1+v_2$ which is possible.