Realistically speaking, the 10kg ball could not transfer all of its momentum to the 1 kg ball: the mass ratio is far too big. Instead, the 10 kg ball will keep on rolling forward at something just under its initial velocity and the 1 kg ball will be sent forward at about 2*v₁.

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Here is the information relevant to the question.

CoM says that (duh) momentum is conserved - it doesn't say that a 1kg ball can stop a 10kg ball dead in its tracks. Balls of equal mass exchange velocities in that way; unequal masses don't.

Running via wiki, I find a pair of equations (under 'Elastic collisions') allowing you to determine the final velocity, v(1,2), from the initial velocities u(1,2) and the masses m(1,2). For your 10kg mass, the equation reads:

v1 = ((m1-m2)/(m1+m2))*u1 + ((2*m2)/(m1+m2))*u2

Given that u2 (initial velocity of the smaller ball) = 0, and inputting the other figures, that gives:

v1 = (9/11) * 1 = ~0.82 m/s.
k.e. = 1/2 m1*v1^2 = 3.35 J.

The equation for m2 is the same, except with the subscripts flipped. Running the numbers gives us:

v2 = (20/11) * 1 = ~ 1.82 m/s
k.e. = 1.65 J.

And, of course,

3.35 + 1.65 = 5 J.

Does that answer your question?

(Short version: a bowling ball rolled at a ping-pong ball won't fire the ping-pong ball off into orbit)

hS