If a ball is thrown to a person standing on a frictionless surface, is the impulse of the thrower equal to that of the catcher? If a person throws a ball, exerting a given impulse does the person that catches the ball receive the same impulse assuming that the catcher moves. Is the impulse that the catcher receives less than the impulse that the thrower receives because the ball continues to move with the catcher or does the catcher receive all of the impulse, to begin with and then return momentum to the ball as they pull it along in their hand?
 A: The change in the ball's momentum when caught will be smaller than when thrown, since its speed will not go all the way back to zero. Since $J=\Delta p$, this means the impulse imparted by the ball on the catcher will be smaller than the impulse imparted on the ball by the thrower. This is why, if (say) a baby falls from a balcony above, you would want to move your arms downward as you catch it, instead of simply holding your arms out.
Momentum within the system is of course conserved, but this just means that the ball exerted an equally big impulse back on the thrower as the thrower exerted on it. Also, and the catcher exerted an equally small impulse on the ball as the ball exerted on them. I hope this makes sense.
A: Momentum is conserved in any inertial frame. You choose an inertial frame to write the equations for momentum conservation. In the catcher's frame obviously the ball's momentum is less than that in thrower's frame due to relatively slow motion.
A: Let both the men have a mass of M and the ball have a mass of m. Now, let's say the first man throws the ball at an angle $\theta$ and velocity v towards the second man.

Now, just after he has thrown the ball we will see this:

The velocity of the first man will be $\frac{mv \cos{\theta}}{M}$ in the opposite direction of the ball's motion by conservation of momentum in the horizontal direction, as no forces act in that direction.
Now, just before the ball reaches the other man this will be the situation:

Since the man and the ball move together after he catches the ball, we can treat it like an inelastic collision, giving us a velocity of the combined ball-man system as $\frac{mv\cos{\theta}}{M+m}$,

Here, we can clearly see that the final velocity of the first man$\left(\frac{mv \cos{\theta}}{M}\right)$ is more than that of the second man-ball system$\left(\frac{mv\cos{\theta}}{M+m}\right)$ and their masses are equal. So, we can say that the impulse given to the second man is lesser than that given to the first man.
Hope this helps! 
P.S.: Sorry for the bad drawings
A: As we know from the definition of impulse that $$ J = \int \vec{F} dt$$ and thus as thrower exert a Force for short time while the catcher have longer time for taking a force , So we concluded that Thrower have greater impulse than catcher. But there is also $\vec{F}$ sitting there,  To make it simple let us suppose the thrower throws a ball vertically upward with a force say $\vec{F_1}$ for a short interval $dt$. While force exert by ball $mg$ on catcher,that is constant for short interval say $\Delta t$. Now  which one is greater?  Let's take the limit that the thrower exert a really high force in that case force exert by ball will be same as $mg$.Thus we concluded that Impulse by thrower is greater that catcher in all the cases. 
