Question

Two identical balls A and B having velocities of 0.5 m/s and -0.3 m/s respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be
A. 0.3 m/s and 0.5 m/s
B. -0.5 m/s and 0.3 m/s
C. 0.5 m/s and -0.3 m/s
D. -0.3 m/s and 0.5 m/s

Answer 1

Both the balls given are identical and the collision is completely elastic. So, the momentum will be conserved as it always is during collisions and as the collision is elastic, kinetic energy will also remain conserved as no energy is lost. We will get the result that the velocities of both the balls get exchanged.

Complete step by step answer:
First let us assume the mass of the balls to be m and velocity of ball A to be vA and velocity of ball B to be $v_B$ after the collision. Law of conservation of momentum always holds so first, we will use that.
$\begin{aligned} & m\times 0.5+m\times (-0.3)=m\times {{v}_{A}}+m\times {{v}_{B}} \\\ & \Rightarrow 0.5-0.3=0.2={{v}_{A}}+{{v}_{B}} \\\ \end{aligned}$
As it is given that the collision is elastic, there will be no loss of energy and the total kinetic energy of both the balls will be the same.
$\begin{aligned} & \dfrac{m{{(0.5)}^{2}}}{2}+\dfrac{m{{(-0.3)}^{2}}}{2}=\dfrac{m{{({{v}_{A}})}^{2}}}{2}+\dfrac{m{{({{v}_{B}})}^{2}}}{2} \\\ & \Rightarrow 0.25+0.09=0.34={{v}_{A}}^{2}+{{v}_{B}}^{2} \\\ \end{aligned}$
Now, we have two equations and two variables so they can be solved.

& {{({{v}_{A}}+{{v}_{B}})}^{2}}={{v}_{A}}^{2}+{{v}_{B}}^{2}+2{{v}_{A}}{{v}_{B}}={{0.2}^{2}}=0.04 \\\ & {{v}_{A}}^{2}+{{v}_{B}}^{2}=0.34 \\\ & 2{{v}_{A}}{{v}_{B}}=0.04-0.34=-0.3 \\\ & {{v}_{A}}=\dfrac{-0.15}{{{v}_{B}}} \\\ & {{v}_{B}}+\dfrac{-0.15}{{{v}_{B}}}=0.2 \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow {{v}_{B}}^{2}-0.2{{v}_{B}}-0.15=0 \\\ & {{v}_{B}}=\dfrac{-(-0.2)\pm \sqrt{0.04-4\times 1\times (-0.15)}}{2}=\dfrac{0.2\pm \sqrt{0.64}}{2}=\dfrac{0.2\pm 0.8}{2} \\\ \end{aligned}$$ $v_B$ can have two different values. $$\begin{aligned} & {{v}_{B}}=0.5 \\\ & {{v}_{B}}=-0.3 \\\ \end{aligned}$$ The first value will be taken as the ball could not continue on their path so their velocities must have changed. Both the balls will acquire the velocity of the other one after the collision. **Hence, the correct option is D, i.e. -0.3 m/s and 0.5 m/s.** **Note:** Whenever two identical object collide and their collision is elastic, they both acquire the velocity of the other because when we solve the two equations that we get from law of conservation of momentum and law of conservation of kinetic energy the two variables are identical and will form a quadratic equation and have the same solution which will be the values of their velocities before collision and as they cannot continue on their path, the value of velocity they had before collision is disregarded as the solution and what remains is the velocity of the other one.