2019-03-01_Physics_Times

So momentum of two pieces after explosion

( cos )

2 2

m m

  v  V

By the law of conservation of momentum

cos cos 3 cos.

2 2

m m

mv  v V V v 



   

2

2

p

E

m

 ( Linear momentum is conserved )

1 2

2 1

1 E m

E

m E m

   .

As momentum is conserved the momentum of

both the particles are equal to P.

Energy of explosion  K E K E. 1. 2

2 2 2

1 2

1 2 1 2

( )

2 2 2

P P P m m

m m mm



  

The velocity of the canon at the heightest point

is vcos . Let m is the mass of the canon.

P Pi f

vcos (0) v'

2 2

m m

m  

v 2vcos' 

The y position of CM must be zero.

y = -5m

Let

3

15

4 4

cm^0

m m

y

y

m

 

  

   

2

0 0

1

2

E KE mv 

Where v 0 is the velocity of one of three fragments

From conservation of LM

0 0 0

0    mv m v i v j v k ˆ ˆ ˆ

 



8.Sol:

9.Sol:

10.Sol:

11.Sol:

12.Sol:

v v (^30)
The energy of explosion is
 
(^22)
0 0
1 1
3 3
2 2
KE m v mv    
 
2
0
1
6
E mv  6 E 0