Two balls, A and B, move towards each other. Ball A has a change in momentum of 2.4 kg·m·s⁻¹ to the left. Ball B has a mass of 0.3 kg and an initial velocity of 15 m·s⁻¹ to the left. What is the initi

Let's choose RIGHT as the positive direction.

• Mass of ball B, $m_B = 0.3 \text{ kg}$
• Initial speed of ball B, $v_{iB} = 15 \text{ m}\cdot\text{s}^{-1}$. Since ball B moves towards ball A (which is initially moving right), ball B's initial velocity is to the LEFT. So, $v_{iB} = -15 \text{ m}\cdot\text{s}^{-1}$.
• Change in momentum of ball A, $\Delta p_A = 2.4 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$ to the LEFT. So, $\Delta p_A = -2.4 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$.

According to Newton's Third Law, the impulse exerted by ball A on ball B is equal in magnitude and opposite in direction to the impulse exerted by ball B on ball A. Therefore, the change in momentum of ball B ($\Delta p_B$) is equal in magnitude and opposite in direction to the change in momentum of ball A ($\Delta p_A$).

$$\Delta p_B = -\Delta p_A$$ $$\Delta p_B = -(-2.4 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1})$$ $$\Delta p_B = +2.4 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$$

This means the change in momentum of ball B is $2.4 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1}$ to the RIGHT.

The change in momentum for ball B is given by:

$$\Delta p_B = m_B v_{fB} - m_B v_{iB}$$

Substitute the known values:

$$2.4 \text{ kg}\cdot\text{m}\cdot\text{s}^{-1} = (0.3 \text{ kg}) v_{fB} - (0.3 \text{ kg})(-15 \text{ m}\cdot\text{s}^{-1})$$ $$2.4 = 0.3 v_{fB} + 4.5$$

$$0.3 v_{fB} = 2.4 - 4.5$$ $$0.3 v_{fB} = -2.1$$ $$v_{fB} = \frac{-2.1}{0.3}$$ $$v_{fB} = -7 \text{ m}\cdot\text{s}^{-1}$$

The negative sign indicates that the final velocity of ball B is to the LEFT.

The velocity of ball B after the collision is $\boxed{\text{7 m}\cdot\text{s}^{-1} \text{ to the LEFT}}$.