# 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?

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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` .**

**Sources:**

known data:

m=0.1kg,h=12m

unknown data:

momentum=?

formula:

momentum=mass*velocity

velocity(V^2)=Vo^2+2ax

Vo=initial displacement(i.e)=0

a=gravitational force(m/s)=9.81m/s

h (or) x=displacement(m)

momentum=0.1*(2*9.8*12)^1/2

=1.5 kg m/s.