Question

A ball of mass 1 kg is dropped from 20 m height on ground and it rebounds to height 5m. Find magnitude of change in momentum during its collision with the ground.[ Take g =10m/s^{​​​​​2​​}

Answer 1

For downward journey, the final velocity just before striking the ground is given by,
$v^2-u^2=2as$
$v^2=2\times 10\times 20=400$

$v=20$ m/s
Therefore initial momentum = mu = 1 x 20 = 20 kg-m/s
After striking the ground for upward motion the initial velocity is given by,
$v^2-u^2=2as$
$0-u^2=2\times (-10)\times 5$
$u=10$ m/s.
Final Momentum = mv = 1 x 10 = 10 kg-m/s
Therefore, the magnitude of the change in momentum = mv-mu = 10- (-20) = 30 kg-m/s.