Question

Ball A 1.55kg moving right at 8.76 m/s makes a head-on collision with ball B (0.752 kg) moving left at 11.4 m/s. After, ball B moves right at 9.03 m/s. What is the final velocity of ball A?

Answer 1

1.15 m/s to the left (3 sig. fig.).

Explanation

Momentum is conserved between the two balls if they are not in contact with any other object. In other words,

$p_{\text{A,initial}} + p_{\text{B,initial}}=p_{\text{A,final}} + p_{\text{B,final}}$

$m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}}$, where

• $m$ stands for mass and
• $v$ stands for velocity, which can take negative values.

Let the velocity of objects moving to the right be positive.

• $m_\text{A} = 1.55\;\text{kg}$,
• $m_\text{B} = 0.752\;\text{kg}$.

Before the two balls collide:

• $v_\text{A} = +8.76\;\text{m}\cdot\text{s}^{-1}$,
• $v_\text{B} = -11.4\;\text{m}\cdot\text{s}^{-1}$.

After the two balls collide:

• $v_\text{A}$ needs to be found,
• $v_\text{B} = +9.03\;\text{m}\cdot\text{s}^{-1}$.

Again,

$m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}}$,

$1.55 \times (+8.76) + 0.752 \times (-11.4) = 1.55\;{\bf v_{\textbf{A,final}}} + 0.752 \times (+9.03)$.

$v_{\text{A,final}} = \dfrac{1.55 \times (+8.76) + 0.752 \times (-11.4)-0.752 \times (+9.03)}{1.55} = -1.15\;\text{m}\cdot\text{s}^{-1}$.

$v_{\text{A,final}}$ is negative? Don't panic. Recall that velocities to the right is considered positive. Accordingly, negative velocities are directed to the left.

Hence, ball A will be travelling to the left at 1.15 m/s (3 sig. fig. as in the question) after the collision.

Answer 2

10.5 m/s. Not sure if it’s correct.