Grade 12th passMechanics

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

anshu

8 Years agoGrade 12th pass

4 Answers

sagar kumar

8 Years ago

To solve, the problem

first, of all

use the 3^rd 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......

Khushi

8 Years ago

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

pooja kapgate

8 Years ago

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

Yash Chourasiya

6 Years ago

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