Question

Two billiard balls with the same mass m move towards one another. Ball one travels in the positive x-direction with a speed of v1i, and ball two travels in the negative x-direction with a speed of v2i. The two balls collide elastically, and both balls change direction after the collision. If the initial speeds of the balls were v1i=2.0m/s and v2i =1.0m/s, what would be the final speed and direction of ball two, v2f, in m/s?

Answer 1

Answer:

$v_{2f}=2m/s$ in the positive x-direction.

Explanation:

Conservation of momentum tells us that the momentum before the collision must be equal to the momentum after the collision: $p_i=p_f$.

Before the collision we have:

$p_i=m_1v_{1i}+m_2v_{2i}=mv_{1i}+mv_{2i}=m(v_{1i}+v_{2i})$

And after the collision:

$p_f=m_1v_{1f}+m_2v_{2f}=mv_{1f}+mv_{2f}=m(v_{1f}+v_{2f})$

Since they must be equal we have:

$m(v_{1i}+v_{2i})=m(v_{1f}+v_{2f})$

Which is the same as:

$v_{1i}+v_{2i}=v_{1f}+v_{2f}$

And can be written as:

$v_{1i}-v_{1f}=v_{2f}-v_{2i}$

Since the collision is elastic, kinetic energy is conserved: $K_i=K_f$

Before the collision we have:

$K_i=\frac{m_1v_{1i}^2}{2}+\frac{m_2v_{2i}^2}{2}=\frac{mv_{1i}^2}{2}+\frac{mv_{2i}^2}{2}=\frac{m}{2}(v_{1i}^2+v_{2i}^2)$

And after the collision:

$K_f=\frac{m_1v_{1f}^2}{2}+\frac{m_2v_{2f}^2}{2}=\frac{mv_{1f}^2}{2}+\frac{mv_{2f}^2}{2}=\frac{m}{2}(v_{1f}^2+v_{2f}^2)$

Since they must be equal we have:

$\frac{m}{2}(v_{1i}^2+v_{2i}^2)=\frac{m}{2}(v_{1f}^2+v_{2f}^2)$

Which is the same as:

$v_{1i}^2+v_{2i}^2=v_{1f}^2+v_{2f}^2$

And can be written as:

$v_{1i}^2-v_{1f}^2=v_{2f}^2-v_{2i}^2$

Which is the same as:

$(v_{1i}-v_{1f})(v_{1i}+v_{1f})=(v_{2f}-v_{2i})(v_{2f}+v_{2i})$

But since we already proved that $v_{1i}-v_{1f}=v_{2f}-v_{2i}$, we can cancel those terms and we have that:

$v_{1i}+v_{1f}=v_{2f}+v_{2i}$

So we have the system:

$v_{1i}-v_{1f}=v_{2f}-v_{2i}$

$v_{1i}+v_{1f}=v_{2f}+v_{2i}$

Adding both sides between them we obtain:

$v_{1i}-v_{1f}+v_{1i}+v_{1f}=v_{2f}-v_{2i}+v_{2f}+v_{2i}$

Which is:

$2v_{1i}=2v_{2f}$

Which means:

$v_{2f}=v_{1i}=2m/s$

Where we used a positive velocity since ball 1 travels in the positive x-direction at the beginning.