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Grade: 12th pass

                        

A ball of mass 1kg is dropped from 20m height on ground and it rebounds to height 5m. Find magnitude of change in momentum during its collision with the ground

3 years ago

Answers : (4)

sagar kumar
99 Points
							
To solve, the problem
first, of all
use the 3rd equation to find the value of v (where put the value of u=o, s=20m, g=9.8ms-2) you will get the value of v
after that, again use the third equation to find the u while rebounding (where put the value of v=0 ,s=5m,g=9.8m/s2) you will get the value of u thereafter you must be knowing that, Impulse = change in momentum
then put , mv-(mu) =mv+mu=m(v+u) then put the values of m,v,u you will get the magnitude and change in momentum......
 
 
3 years ago
Khushi
19 Points
							
for downward motion ;
u=0 m/s
a=10 m/s2
s=20 m
using v2- u2=2as
v= 20m/s
for upward motion;
v=0 m/s
a=10 m/s2
s=5 m using v2 – u2=2as
u= -10 m/s
I=mv – mu
I=20+10=30 kg m/s
3 years ago
pooja kapgate
21 Points
							1stly consider coefficient of restitution e^1/2 then apply          v^2=2gh           =2×10×20      =400  V=20m/s Then v`/v=e^1/2           If v =20 then v`=10 so  V`+v=20+10=30m/sNow , p=mv              =30×1=30 kg m/s
						
2 years ago
Yash Chourasiya
askIITians Faculty
246 Points
							Dear Student

A ball of mass 1 kg dropped from 20 m height on ground and it rebounds to height 5.

Using energy conservation theorem:
mgh = ½(mv2)
v = √2gh
where, g = 10 m/s² and h = 20 m
v = √2*10*20
v = 20 m/sec

Momentum before collision = mv
= 1 x 20
= 20 kgm/s

After rebounding the ball.
Using energy conservation theorem:
mgh = ½(mv2)
v = √2gh
where, g = 10 m/s² and h = 5 m
v = – √2*10*5
v = – 10 m/sec

Momentum after collision = mv
= -1 x 10
= -10 kgm/s

Change in momentum during collision = After collision - Before collision
= 20 - (-10)
= 30 kgm/sec

Hence, The change in momentum is 30 kgm/sec


I hope this answer will help you.
Thanks & Regards
Yash Chourasiya
3 months ago
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