Question

a 2 kg ball moving at 4 m/s collides elastically with a 3 kg ball at rest. what is the velocity of the 2 kg ball after the collision, if the final velocity of the 3 kg ball is 2 m/s? $m_1v_{i1} + m_2v_{i2} = m_1v_{f1} + m_2v_{f2}$
options: -0.4 m/s, 0 m/s, 2.4 m/s, 1 m/s

Explanation:

Step1: List given values

$m_1=2\ \text{kg}$, $v_{i1}=4\ \text{m/s}$, $m_2=3\ \text{kg}$, $v_{i2}=0\ \text{m/s}$, $v_{f2}=2\ \text{m/s}$

Step2: Rearrange momentum equation

Solve $m_1v_{i1} + m_2v_{i2} = m_1v_{f1} + m_2v_{f2}$ for $v_{f1}$:
$v_{f1} = \frac{m_1v_{i1} + m_2v_{i2} - m_2v_{f2}}{m_1}$

Step3: Substitute values into formula

$v_{f1} = \frac{(2\times4) + (3\times0) - (3\times2)}{2}$

Step4: Calculate final velocity

$v_{f1} = \frac{8 + 0 - 6}{2} = \frac{2}{2} = 1\ \text{m/s}$

Answer:

1 m/s