Intuitive understanding of momentum problem

[yellowjellybean]

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0.5-kg ball traveling at 10 m/s collides with a stationary 2.0-kg ball and rebounds in the opposite direction at 6 m/s. What is true of the speed of the 2.0- kg ball after collision?
A. The2.0-kgballmovesat10m/s.
B. The2.0-kgballmovesat6m/s.
C. The2.0-kgballmovesat4m/s.
D. The2.0-kgballmovesat2.5m/s.


Solution
This is a collision like the one seen in Case 5, where the impact ball rebounds off of an initially stationary ball. Intuitively we know that the 2.0-kg ball will be moving faster than 2.5 m/s, the speed it would have attained had the impact ball come to rest. Because the impact ball rebounds, the 2.0-kg ball gets all of the transferred momentum plus some recoil momentum. The recoil momentum is small, its exit speed should be just a little greater than 2.5 m/s, making choice C the most probable answer. To solve precisely, we can apply the following math:

I don't understand the bolded statement above. INTUITIVELY I don't understand why the velocity of the 2.0 kg ball would be higher if the 0.5kg ball recoils than if the 0.5kg ball comes to rest. Mathematically it totally makes sense.

 

[mehc012]

When it doesn't rebound, ball A has to push hard enough to come to a stop.
When it does rebound, ball A pushes hard enough to come to a stop, and then pushes off to start moving back in the opposite direction.

 

[wjs010]

i hope im doing this right because im doing mcat study as well and havent covered all of physics yet. As i remember from physics 1, mv1i + mv2i = mv1f + mv2f. since 2 isn't moving initially, mv2i is 0. since 1 bounces back in the opposite direction ( and velocity is a vector, not scalar) at -6m/s, the equation comes out like this: (.5)(10)=(.5)(-6) + 2v

5 = -3 + 2v
8=2v
4=v

in this situation, I wouldn't even try to worry about the intuitive part..the math can be done in your head!

edit: damn i just saw you know how to do the math part..oh well lol