A man throws a ball of mass 0.5 kg vertically upward with a velocity of 25 m/s. Find :

(a) the initial momentum of the ball

(b) momentum of the ball at the half way mark of the maximum height (given g = 10 m/s2 )
A man throws a ball of mass 0.5 kg vertically upward with a velocity of 25 m/s. Find :

(a) the initial momentum of the ball

(b) momentum of the ball at the half way mark of the maximum height (given g = 10 m/s2 )
(a)
initial momentum = mass x initial velocity
or
p_{i} = m x u
here,
m = 0.5 kg
u = 25 m/s
so,
p_{i} = 0.5 x 25
thus,
initial momentum of ball will be
p_{i} = 12.5 kg.m/s
.
(b)
we know that
v^{2}  u^{2} = 2as
or
s = (v^{2}  u^{2}) / 2a
here,
final velocity (at max. height), v = 0
u = 25 m/s
a = g = 10 m/s2
so,
s = (0  25^{2}) / 2x10
thus, maximum height attained
s = 31.25m
so,
velocity at midpoint or s/2 = 31.25/2 = 15.625m will be
v'^{2} = u^{2} + 2a(s/2)
or
v'^{2} = 25^{2} + 2x10x(15.625)
or
v'^{2} = 312.5
thus, velocity at halfway mark
v' = 17.67 m/s
so,
momentum at halfway mark
p' = m x v' = 0.5 x 17.67
thus,
p' = 8.835 kg.m/s