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please solve my activity urgently then I can mark you as brainallist.... and don't spam answer..... otherwise I will report.......
PLZZ solve my problem......
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Answer:
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Answer:
To prove:
The realtive velocity of approach is equals to the realative velocity of sapration.
To derive :
Pf= Pi
Derivation:
in elastic collision , linear momnetum and energy both are conserved .
let A body of mass m moving with u and B body of mass M moving with U velocity . after elastic collision their velocity v and V respectively .
use law of conservation of linear momentum ,
Fext = 0
so,
Pi = Pf
mu + MU = mv + MV
m( u -v) = -M(U - V) ------------(1)
now,
from conservation of energy ,
KEi = KEf
1/2mu² + 1/2MU² = 1/2mv² + 1/2MV²
1/2m(u² -v²) = 1/2M(V² -U²)
m(u² -v²) = - M(U²- V²)
m(u -v)(u + v) = - M( U - V)( U + V)
from equation (1)
(u + v) = ( U + V)
u - U = V - v
1 = ( V - V)/(u - U)
-1 = (V - v)/(U - u) =(v - V)/(u -U)
here ( v-V)/( u - U) is known as coefficient of restitution, where ( v-V) velocity of seperation and (u -U) is velocity of approach .
according to coefficient of restitution
- e = (v - V)/(u - U)
compare above to this ,
-e = -1
e = 1
\bold{\huge{\fbox{\color{blue}{Hence, verified}}}}
Hence, verified