To determine the velocity of 2 g marble after the collision, we use the equation for conservation of momentum.
Momentum before the collision= Momentum after the collision.
\(m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}\).
Here, \(m_{1}\) is the mass of larger marble, \(m_{2}\) is the mass of smaller marble, \(u_{1}\) is the velocity of the larger marble and \(u_{2}\) is the velocity of smaller marble before collision and \(v_{1}\) is the velocity of the larger marble and \(v_{2}\) is the velocity of smaller marble after collision.
Given, \(m_{1} = 2g\),\(m_{2} = 1.25\ g\),\(u_{1} =0\ m /s\),\(u_{2} =8\ m/s\), \(v_{1} =?\) and \(v_{2} =1\ m/s\).
Substituting these values in above equation,
\(2g\times 0 + 1.25g\times 8\ m/s = 2g\times v_{1} + 1.25g\times\ 1 m/s\\\\ v_{1}=\frac{10-1.25}{2} =4.375\simeq 4m/s.\)
Thus, the velocity of 2 g marble after the collision is \(4m/s\)
