a ball with a mass of 0.10 kg is dropped from 12 m high. what is the ball's momentum when it hits the ground?

Answer 1

The object's speed can be found by using the formula v = sqrt(2gy) = sqrt(2*9.8*12) = 15.3 m/s. (The formula is derived from the energy balance mgy = 0.5*mv^2, where mass cacels out.) Momentum is just mv = 0.1*15.3 = 1.53 kg-m/s.

Answer 2

Momentum P = m*v (1)
where m is the mass of the ball and v its velocity.
Since kinetic energy is K=(1/2)*m*v^2 (2)
we can solve (1) for v
v=P/m
and combine with (2) to get
K=P^2/(2*m) (3)

Let's assume that the ball is dropped from point A (at height h=12m) and reaches the ground at point B.
Applying the law of conservation of mechanical energy we get
K(A) + U(A) = K(B) + U(B)
Since the ball is dropped (and not THROWN) its initial kinetic energy equals zero.
On the other hand U(B)=0 since B is on the ground.
So the above equation gives us
U(A) = K(B)

Combining with (3) we get
U(A) = P(B)^2/(2*m) ->
m*g*h = P(B)^2/(2*m) ->
P(B) = SQRT[ 2*m^2*g*h ]
where SQRT is the square root.
Using the values given we get
P(B) = 1.55 kgr*m/s
(approximate value, assuming that g=10m/s^2)

Answer 3

Ek=1/2 * m * v^2
p=m*v
where,
where p is the momentum, m is the mass, and v the velocity.

v=sqrt(2*g*h)
=sqrt(2*9.81*12)
=sqrt(235.44)
=15.344 meter/second

Now, p= m*v
=0.1*15.344
=1.5344 kg-m/second