a marble of 100g dropped from the top of a tower of height 5m and comes to rest when it hits the ground the change in momentum of thr marbel in kg ms is??
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 i
Answer:
impulse=FΔt=mΔv
so we have to find the velocity with which ball strikes the ground : v
1
=
2gh
=
2×10×20
=20
s
m
and for finding speed with which it rebounds such that it goes to a height of 5m will be
v
2
=
2gh
=
2×10×5
=10
s
m
so, impulse=m(v
2
−v
1
)=0.10×(20−10)=1N−s
F=
dt
dP
=1/0.01=100N
Hope it HELPS you