Question

An iron sphere of mass 1kg id dropped from a height of 10m . if the acceleration of the sphere is 9.8m/second. Calculate the momentum transferred from the ground by the ball

Answer 1

momentum=mv

given h=10m

a=9.8=g

m=1kg

u=0

v^2-u^2=2as

v^2=2*9.8*10

v^2=196

v=14

momentum P=mv

=1*14

=14 kgms^-1

Answer 2

Hello

Answer

HERE

Initial velocity (u) = 0

Distance travelled (s) = 10 m

Acceleration of sphere (a) = $9.8 ms^{-2}$

Final velocity of sphere when it just reaches the ground can be calculated

using The formula

= $v^2$ - $u^2$ =  $2as$

= $v^2$ - $0$ = $2$ * $9.8ms^{-2}$* $10m$

= $196m^2s^{-2}$

= $v = \sqrt{196m^2s^{-2}$

= $14ms^{-1}$

Momentum of the sphere just before it touches the ground = mv

= $1kg * 14ms^{-1} = 14kgms^{-1}$

On reaching the ground , the iron sphere comes to rest , so its final momentum is equal to zero

NOW

According to the law of conservation of momentum

Momentum transferred to the ground = momentum of the sphere just before it comes to rest

= $14kgms^{-1}$

                                                                                                                                     

Hope it helps you