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a body of mass 10kg collides elastically with a stationery body of mass 4kg,the initial k.e. of the first body ise,the amount of k.e. left over with the first body is

a body of mass 10kg collides elastically with a stationery body of mass 4kg,the initial k.e. of the first body ise,the amount of k.e. left over with the first body is

Grade:11

1 Answers

Anil Pannikar AskiitiansExpert-IITB
85 Points
13 years ago

Dear Chakrala ,

 

conserve momentum,

10*V1 = 10V2 +4*V'1   ........................1

given initial KE ,  e = 10 *V12/2    ....................2

energy conservation ,

e  = 10*V22/2 + 4*V'12/2   ..............................3

For elastic collision, velocity of approach equals velocity of separation. Hence,

V1 - 0 = V'1 - V2   .............................4

use the 4 equation to find value of  V1, V2, V'1 and e

Left over KE of first body =  KE final = 10*V22/2

 

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Askiitians Expert

Anil Pannikar

IIT Bombay

 

 

 

For elastic collision, velocity of approach equals velocity of separation. Hence,



 
    
      v
      
        1
        i
      
    
    −
    
      v
      
        2
        i
      
    
    =
    
      v
      
        2
        f
      
    
    −
    
      v
      
        1
        f

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