Answer to Question #256545 in Physics for megs

Question

During a trick shot demonstration, a pool player shoots a 170 g cue ball directly at a 150 g billiard ball that is moving towards the cue ball with a velocity of 1.2 m/s. The perfectly elastic collision between the billiard ball and the cue ball occurs when the cue ball is moving at a velocity of 5.9 m/s. Determine the velocity of the billiard ball and the cue ball directly after the collision.

Note: For this question, please include all of your work on your written solution. Include the velocity for both the cue ball and the billiard ball. There is no need to place your answer in the answer box for this problem (it only checks the velocity of the cue ball).

Answer 1

We can find the velocity of the billiard ball and the cue ball directly after the collision from the law of conservation of momentum. From the law of conservation of momentum, we have:

$$m_1u_1+m_2u_2=m_1v_1+m_2v_2. (1)$$

Since collision is perfectly elastic, kinetic energy is conserved and we can write:

$$\dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2. (2)$$

Let’s rearrange equations (1) and (2):

$$m_1(u_1-v_1)=m_2(v_2-u_2), (3)$$ $$m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2). (4)$$

Let’s divide equation (4) by equation (3):

$$\dfrac{(u_1-v_1)(u_1+v_1)}{u_1-v_1}=\dfrac{(v_2-u_2)(v_2+u_2)}{v_2-u_2},$$ $$u_1+v_1=u_2+v_2. (5)$$

Let's express $v_2$ from the equation (5) in terms of $u_1$, $u_2$ and $v_1$:

$$v_2=u_1-u_2+v_1. (6)$$

Let’s substitute equation (6) into equation (3). After simplification, we get:

$$(m_1-m_2)u_1+2m_2u_2=(m_1+m_2)v_1.$$

From this equation we can find $v_1$:

$$v_1=\dfrac{(m_1-m_2)}{(m_1+m_2)}u_1+\dfrac{2m_2}{(m_1+m_2)}u_2,$$ $$v_1=\dfrac{(0.17\ kg-0.15\ kg)}{(0.17\ kg+0.15\ kg)}\times5.9\ \dfrac{m}{s}+\dfrac{2\times0.15\ kg}{(0.17\ kg+0.15\ kg)}\times(-1.2\ \dfrac{m}{s}),$$ $$v_1=-0.76\ \dfrac{m}{s}.$$

The sign minus means that the que ball moves in the opposite direction (to the left) after the collision.

Substituting $v_1$into the equation (6) we can find $v_2$:

$$v_2=\dfrac{2m_1}{(m_1+m_2)}u_1+\dfrac{(m_2-m_1)}{(m_1+m_2)}u_2,$$ $$v_2=\dfrac{2\times0.17\ kg}{(0.17\ kg+0.15\ kg)}\times5.9\ \dfrac{m}{s}+\dfrac{(0.15\ kg-0.17\ kg)}{(0.17\ kg+0.15\ kg)}\times(-1.2\ \dfrac{m}{s}),$$ $$v_2=6.34\ \dfrac{m}{s}.$$

The sign plus means that the billiard ball moves in the opposite direction (to the right) after the collision.