Question

a 2.5 kg ball moving at 4.2 m/s collides with a 2.5 kg stationary ball in a perfectly elastic collision. what is the velocity of the second ball after the collision?

a. 4.2 m/s
b. 2.5 m/s
c. 8.4 m/s
d. 0 m/s

Explanation:

Step1: Apply conservation of momentum

For a perfectly elastic collision between two objects of equal mass (\(m_1 = m_2=2.5\space kg\)), the conservation of momentum equation is \(m_1u_1 + m_2u_2=m_1v_1 + m_2v_2\). Given \(u_2 = 0\space m/s\) (second ball is stationary), the equation simplifies to \(m_1u_1=m_1v_1 + m_2v_2\). Since \(m_1 = m_2\), we have \(u_1=v_1 + v_2\).

Step2: Apply conservation of kinetic energy (for elastic collision)

The conservation of kinetic energy equation is \(\frac{1}{2}m_1u_1^{2}+\frac{1}{2}m_2u_2^{2}=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}\). With \(u_2 = 0\space m/s\) and \(m_1 = m_2\), we get \(u_1^{2}=v_1^{2}+v_2^{2}\).

Step3: Solve the system of equations

From \(u_1=v_1 + v_2\), we have \(v_1=u_1 - v_2\). Substitute into \(u_1^{2}=v_1^{2}+v_2^{2}\):

$$
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Given \(u_1 = 4.2\space m/s\), so \(v_2 = 4.2\space m/s\)

Answer:

A. \(4.2\space m/s\)