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?
2 Answers
sciencesolve | Teacher | (Level 3) Educator Emeritus
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` .
vickyprabhucool | Student, Undergraduate | eNotes Newbie
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.