Find the momentum of an object of mass 2.0kg which falls from rest for 10seconds
Answers
Answer:
mass, m = 1kg
height, s = 20m
initial velocity of ball, u = 0m/s
acceleration, a = 10m/s
2
Using, v
2
= u
2
+ 2as
= 0 + 2×10 ×20
= 400
v = 20m/s
Now, let us take upward direction as positive and downward direction as negative.
Initial momentum of the ball (before striking the ground) = m×v
= 1×(-20) (velocity is downward, hence negative)
= -20kgm/s
Now, when the ball strikes and rebounds, its velocity becomes +20m/s (after striking, it moves upward, hence positive velocity)
Final momentum (after striking the ground) = m×v
= 1×(+20)
= +20kgm/s
Change in momentum = Final momentum - Initial momentum
= 20kgm/s - (-20kgm/s)
= 40kgm/s in upward direction.
Answer:
When it’s still at the height of 10m, the object has potential energy. The formula for this is Mass x Gravitational Acceleration x Height. (m*g*h)
Mass = 20 kg
Gravitational Acceleration = 9.81 meters/(second^2)
Height = 10m
Multiplying, we get: 20 x 9.81 x 10 x (kg*meters/(second^2)) * meters
Simplifying: Potential energy = 1,962 Newton * meters = 1,962 Joules
If the object falls 10m to the ground, then right before it hits, all of its potential energy will have been converted into kinetic energy (plus a tiny bit of air friction energy lost in heat). So the kinetic energy will be 1,962 Joules *minus* some tiny amount (whatever amount of energy was lost heating up the air on the way down).