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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`
7 years ago

85 Points
```
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

All the best.
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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
```
7 years ago
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