Two identical balls A and B collide head on elastically. If velocities of A and B before the collision are + 0.5 m/s and - 0.3 m/s respectively, then their velocities after the collision will be :

Question

Two identical balls A and B collide head on elastically. If velocities of A and B before the collision are + 0.5 m/s and - 0.3 m/s respectively, then their velocities after the collision will be : (A) + 0.5nysand+ 0.3 m/s (B) - 0.5 m/s and + 0.3 m/s (C) + 0.3 m/s and - 0.5 m/s (D) - 0.3 m/s and + 0.5 m/s

Tardigrade Admin · Accepted Answer

Momentum and energy both are conserved in elastic collision.
Given,

v_{A} = + 0.5 m / s, v_{B} = - 0.3 m / s

Initial momentum = final momentum

m_{A} v_{B} + m_{B} = m_{A} v_{A} + m_{B} v_{B}

\Rightarrow

0.5 + (- 0.3) = v_{A} + v_{B}

?.. (i)
From law of conservation of energy,

\frac{1}{2} (m_{A} v_{A}^{2} + m_{B} v_{B}^{2}) = \frac{1}{2} (m_{A} v_{A}^{2} + m_{B} v_{B}^{2})

?.. (ii)

(0.5)^{2} + (- 0.3)^{2} = v_{A}^{2} + v_{B}^{2}

From Eqs. (i) and (ii),
we get

v_{A} = - 0.3 m / s, v_{B} = + 0.5 m / s

.
Note: In an elastic collision, velocities after collision get interchanged.

Q. Two identical balls A and B collide head on elastically. If velocities of A and B before the collision are +0.5m/s and −0.3m/s respectively, then their velocities after the collision will be :

+ 0.5nysand+ 0.3 m/s

- 0.5 m/s and + 0.3 m/s

+ 0.3 m/s and - 0.5 m/s

- 0.3 m/s and + 0.5 m/s

Q. Two identical balls $A$ and $B$ collide head on elastically. If velocities of $A$ and $B$ before the collision are $+ 0.5 m / s$ and $- 0.3 m / s$ respectively, then their velocities after the collision will be :