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:
r = perpendicular distance from C onto
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)
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.