Projecticle Motion: Clay or Rubber Ball?

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SUMMARY

The discussion centers on the comparison of impulse applied to a door by a 400-gram clay lump versus a 400-gram rubber ball. Both projectiles are thrown with the same speed, but the clay sticks to the door, resulting in a larger time of contact (T) and thus a greater impulse. The impulse is defined as J = F x T, and since the clay has a higher change in momentum due to its sticking nature, it applies a larger impulse than the rubber ball, which bounces off. Therefore, the clay is more effective for closing the door.

PREREQUISITES
  • Understanding of impulse and momentum equations (J = FΔt, J = mΔv)
  • Knowledge of the concept of change in momentum (Δp = P2 - P1)
  • Familiarity with the definitions of mass and velocity in physics
  • Basic grasp of projectile motion principles (though not necessary for this discussion)
NEXT STEPS
  • Study the relationship between impulse and momentum in different collision scenarios
  • Learn about the conservation of momentum in elastic and inelastic collisions
  • Explore practical applications of impulse in real-world physics problems
  • Investigate the effects of varying mass and velocity on impulse and momentum
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Students of physics, educators teaching mechanics, and anyone interested in understanding the principles of impulse and momentum in practical applications.

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You want to close an open door by throwing either a 400-g lump of clay or a 400-g rubber ball toward it. You can throw either object with the same speed, but they are different in that the rubber ball bounces off the door while the clay just sticks to the door. Which projectile will apply the larger impulse to the door and be more likely to close it?
 
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What concepts might be useful here? (You can ignore projectile motion completely.)
 
impulse = F x T
and since they both have equal mass it doesn't matter except

the clay has a larger T value (since it sticks)
so the impulse is larger?

not sure
 
yes, impulse does equal F*t, but it also equals the change in momentum.
 
so the answer is the clay because it has a higher T value = higher momentum?
 
whoa... back up.
First of all: what's the definition of momentum?
Secomdly: are we interested in the momentum of the thrown object or itschange of momentum?

I'd suggest writing down all of the momentum and impulse equations for handy reference.
 
J = FΔt,
J = mΔv
J = P2 - P1
the ones I used/need to know i guess...

anyways I am interested in only the first part.
we are interested in the change of momentum.
so p2-p1 is higher for the clay because it sticks -> higher t.
 
meh i got it so the rubber ball will have a negative velocity while the clay will have equal velocity with the door.
 
to make it more explicit
P2-P1 becomes P2-(-P1) for the rubber ball and P2+P1 so it will have be much higher.
 
  • #10
I think you may have gotten it. State that momentum=mv and \Delta p=m_fv_f-m_iv_i, and I'll be a little more sure.

So in the end of the day, if you're trying to open a door and can't be bothered to get off the sofa, will you chuck the clay or the rubber ball?
 
Last edited:

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