A ball, of mass 0.1 kg, is dropped from a height of 12 m, What is its momentum when it stikes the ground, in kg m/s?
You need to remember the formula of momentum such that:
`p = m*v`
You should use the kinematic equation that helps you to find the velocity such that:
`v^2 = v_0^2 + 2a*x`
Since the initial velocity is zero,the acceleration is g and h represents the height the ball is dropped, you need to substitute these values in the kinematic equation above such that:
`v^2 = 2g*h =gt v = sqrt(2gh)`
Substituting `sqrt(2gh)` for v in equation of momentum yields:
`p = m*sqrt(2gh)`
`p = 0.1*sqrt(2*9.8*12) =gt p = 1.533 kg*m/s`
Hence, evaluating the momentum when the ball strikes the ground yields `p = 1.533 kg*m/s` .