-1

A circus acrobat of mass M leaps straight up with initial velocity $v_0$ from a trampoline. As he rises up, he takes a trained monkey of mass m off a perch at a height h above the trampoline. What is the maximum height attained by the pair?

The only thing I don't understand is why the vertical momentum is conserved when the acrobat catches the monkey despite the presence of gravitational force?

That's the only part in the solution where the conservation of momentum will be used to find the final velocity of acrobat and monkey.

I guessed that it must be because of normal force balancing the gravitational force out but am not sure of it.

tom
  • 6,948
suiz
  • 875
  • This to me looks like a homework type question, but is asking about a clear concept rather than 'check my work' so I am not suggesting close and tried to put a useful answer in, but if someone thinks I have got this wrong then I would be very grateful of a comment to explain. – tom Apr 04 '18 at 11:32
  • 1
    The point is that during the collision the impulse due to the gravitational force is negligible compared with the impulses on the monkeys due to their collision. Related/duplicate? https://physics.stackexchange.com/a/310609/104696 – Farcher Apr 04 '18 at 11:36
  • @sammygerbil - good point - should have spotted that question... – tom Apr 04 '18 at 11:55
  • See also Conservation of Momentum with external force – sammy gerbil Apr 04 '18 at 12:38

1 Answers1

1

Gravity is a distraction here.

The acrobat will be slowing down as (s)he goes up due to gravity, but what we care about is that at the moment the monkey is taken the acrobat and the monkey are together - we combine their masses and use conservation of momentum to find the new upwards velocity.

Often momentum questions are in horizontal planes to avoid gravity, $e.g.$ train on a flat straight track.

Here you need to use gravity to find the velocity at the pick up point, linear momentum conservation to find the new velocity after pick up and then gravity to find how high the pair go with their new velocity.

I thnk the easiest way to understand this is to consider that the pickup takes place in such a sort period of time that the change of velocity due to gravity is negligible.

Hope this helps.

tom
  • 6,948
  • I have added some numbers to the statement you have made in your last paragraph here. https://physics.stackexchange.com/a/310609/104696 – Farcher Apr 04 '18 at 11:42
  • @Farcher - thanks for the reference, didn't realize this was so similar to what you answered on earlier... – tom Apr 04 '18 at 11:56
  • @Farcher isn't the acrobat grabbing a monkey in mid-air inelastic collision, unlike billiard balls that collide elastically? Won't this affect the time of collision? Actually I just kept some values into the result got, I got 45.2m height when gravity was considered and 54.5m while gravity wasn't considered. Isn't this quiet an error or can we let go of it? – suiz Apr 04 '18 at 12:43
  • @suiz yes it is an inelastic collision. Energy will be lost. But the point here is that at the moment of collision you can use momentum conservation to figure out the new combined speed of the acrobat plus the monkey.... – tom Apr 04 '18 at 12:47
  • @tom what about the difference of about 10 m I got between considering gravity and not considering it? Can this error be considered negligible? I guess no. – suiz Apr 04 '18 at 12:55
  • @suiz not sure - but without gravity the acrobat and monkey will go up and up and not stop going up. -- I guess the calculation is that the acrobat starts with v0, but this drops to v1 by height h, then after pickup at same height due to momentum conservation the velocity drops to v2 and then you go up until v=0 – tom Apr 04 '18 at 13:17
  • Yes that how it's done. Maybe to catch the monkey at the same height the indeed needs to do it very fast. Thanks it helped. – suiz Apr 04 '18 at 13:22