Answers

2015-05-18T19:32:18+05:30
After applying horizontal impulse J horizontal velocity of the sphere becomes v 
as initial linear momentum of the system was 0 so
J = Mv 
so v = J/M
now angular impulse about the C.M.
= hMv = Iω  ( I = moment of inertia , ω = angular velocity) 
⇒hMv = 2/5 MR² ×ω
so ω = 5vh/2R²
now while rolling 
v = ωR
⇒v = 5vh/2R
⇒h = 2R/5
0
2015-05-19T13:33:36+05:30

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Sphere of mass M, radius R.  Initial velocity = 0.  Initial angular velocity = 0.
Initial linear and angular momenta = 0.
There is no friction on the surface.  So there is no torque applied by friction.

The horizontal impulse J = Force * Δt = Δp = change in linear momentum
   => J = m v,  where v = velocity of sphere after impulse is given.
     p = m v

  This is according to the conservation of linear momentum
     =>  v = J / m        -- (1)

   Since the line of impulse does not pass through C, center of mass of sphere, there is a torque, angular momentum, or moment of force.  Hence, the sphere rotates.

As the ball starts rolling and does not slip on the smooth surface,
         v = R ω            =>  ω = v / R =  J / (m R)    --- (2)
   where ω = angular velocity of the sphere about the Center of sphere C.

The moment of Inertia I of the solid sphere about its cm C = 2/5 m R²

Angular momentum after impulse = angular impulse given, as L initial =0, by applying the law of conservation of angular momentum:
  
   \vec{L} = \vec{r} X \vec{p}
     r = perpendicular distance from C onto \vec{p} or \vec{J} = h
     L = h * p  = h m v = h m R ω = h m R J / (m R) = h J
         => L = h J      --- (3)

For a solid sphere angular momentum about center C  =  I ω
   =   2/5 m R² ω = 2/5 m R^2 J / (m R)  = 2/5 * R J    --- (4)
 
  compare (3) and (4)
           L = h J = 2/5 * R J
 
   =>  h = 2/5  * R
and we have already, v = J / m        and    ω = J / (m R)
==============
Important info:

   If the impulse is given above the center of mass C and below height  of  2/5 * R, then it moves linearly faster than Rω.    If the impulse is given at a height above 2/5 * R above CM, then the ball has higher ω than v/R,  ie., v < ω R.

  So the ball slips and rolls.

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