0
2.3kviews
A ball of mass 'm' hits directly to a similar ball of mass 'm' which is at rest. The velocity of first ball after impact is zero. Half of the initial kinetic energy is lost in impact
1 Answer
written 7.9 years ago by |
Let A & B be two balls
$UA=u , VA=0\space\space \&\space\space UB=0 , VB=v $
Applying coefficient of restitution eqn,
$e = - (VB - VA)/( uB - uA) = - (v) /(-u) = v/u\text {____(i)} $
Now, initial kinetic energy $= (1/2)mu2 $
Final kinetic energy $= (1/2)mv2 $
But as half the kinetic energy is lost, initial KE= 2* final KE
$(1/2)mu2 = 2* (1/2)mv2 \\ v/u = 1/\sqrt2\text {_____(ii) } $
from (i) & (ii)
$e = 1/\sqrt2 = 0.707 $
Coefficient of restitution is $0.707 $