Grade 11 NEB Practice Set II with solutions | Physics | Physics in Depth
Physics Practice Set II A hose A ball A of mass 0.2 kg, moving with a velocity of 8 m/s, collides directly with a ball B of mass 0.3 kg at rest. Calculate their common velocity if both balls move together. If A had rebounded with a velocity of 3 m/s in the opposite direction after collision, what should be the new velocity of B? Solution : For ball 'A', mass , \(m_1\) = 0.2 kg velocity before collision, \(u_1\) = 8 m/s For ball 'B', mass, \(m_2\) = 0.3 kg velocity before collision, \(u_2\) = 0 Let \(v\) be the common velocity of both balls after collision. From the conservation of linear momentum, \[\begin{align} m_1u_1+m_2u_2&=(m_1+m_2)v\\ 0.2 \times 8 + 0 & = (0.2 + 0.3) v\\ v& = \frac {1.6}{0.5}\\ \therefore v&= 3.2 \hspace{0.1cm} \text{m/s} \end{align}\] If A had rebounded with a velocity of 3 m/s , then, velocity of A...