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# Circular Motion (Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

## DC Pandey (Questions & Solutions) of Physics: NEET

Created by: Ciel Knowledge

## NEET : Circular Motion (Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

``` Page 2

Introductory Exercise 7.3
1. | | v
1
®
= u and | | v
2
®
= v (say)
Dv v v
® ® ®
= + -
2 1
( )
= -
® ®
v v
2 1
D D v u u v v u
® ® ® ® ® ®
× = - × - ( ) ( )
2 1 2 1
= × + ×
® ® ® ®
u v u v
2 2 1 1
- -
® ®
2
2 1
v v
| | | | v v
2
2
1
2 2 2
® ®
+ = + v u
= - + u gL u
2 2
2
| | ( ) Dv
®
= -
2 2
2 u gL
| | ( ) Dv
®
= - 2
2
u gL
2. Ball motion from A to B :
0 2 2
2 2
= + - u g R
min
( )( )
Þ       u gR
min
2
4 =
Ball motion from C to A :
v u g h
2 2
2 = + -
min
( )
= - 4 2 gR gh
Þ v g R h = - 2 2 ( )
3. Decrease in KE of bob
= Increase in PE of bob

L
v
2
â€“ v
1
Di
g
v
1
®
®
®
®
h
v
C
A
2R
B
PE of bob
= K (say)
KE of bob
2
= 1/2 mv
0
v
0
Page 3

Introductory Exercise 7.3
1. | | v
1
®
= u and | | v
2
®
= v (say)
Dv v v
® ® ®
= + -
2 1
( )
= -
® ®
v v
2 1
D D v u u v v u
® ® ® ® ® ®
× = - × - ( ) ( )
2 1 2 1
= × + ×
® ® ® ®
u v u v
2 2 1 1
- -
® ®
2
2 1
v v
| | | | v v
2
2
1
2 2 2
® ®
+ = + v u
= - + u gL u
2 2
2
| | ( ) Dv
®
= -
2 2
2 u gL
| | ( ) Dv
®
= - 2
2
u gL
2. Ball motion from A to B :
0 2 2
2 2
= + - u g R
min
( )( )
Þ       u gR
min
2
4 =
Ball motion from C to A :
v u g h
2 2
2 = + -
min
( )
= - 4 2 gR gh
Þ v g R h = - 2 2 ( )
3. Decrease in KE of bob
= Increase in PE of bob

L
v
2
â€“ v
1
Di
g
v
1
®
®
®
®
h
v
C
A
2R
B
PE of bob
= K (say)
KE of bob
2
= 1/2 mv
0
v
0

1
2
0
2
mv mgh =
v gh
0
2
2 =
v gl
0
2 1 = - ( cos ) q
= ´ ´ ´ - ° 2 5 1 60 9.8 ( cos )
=7 ms
-1

h
PE of bob = K + mgh
```
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