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Jumping on a top pan balance.

by adjacent
Tags: balance, jumping
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adjacent
#1
Sep13-13, 11:56 AM
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A top pan balance measures the force.So if I jump(or fall from a height) on it,Why does it show a greater value?(Acceleration and mass does not change,so the force is same)??
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quawa99
#2
Sep13-13, 12:13 PM
P: 63
When you jump and approach the balance you gain a velocity.When you fall on the pan all of your velocity becomes zero which means your momentum has changed.Applying conservation of momentum the momentum is transferred to the pan and as we know change in momentum over an interval of time causes a force.So then the pan would experience the gravitational force as well as the force caused due to change in your momentum
jfizzix
#3
Sep13-13, 12:17 PM
P: 221
It shows a greater value because the force of impact is in addition to the weight of the object. The weight is the force it takes to keep you standing there motionless. The force of impact is the force it takes to slow you down from your falling speed to a stop on top of the force needed to keep you there.

The larger your momentum before impact, the larger your force of impact, and the greater the value the scale will show.

adjacent
#4
Sep13-13, 12:32 PM
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Jumping on a top pan balance.

Oh,Momentum.Maybe I'll learn it later. :)
quawa99
#5
Sep13-13, 12:44 PM
P: 63
Quote Quote by adjacent View Post
Oh,Momentum.Maybe I'll learn it later. :)
momentum is nothing but M*V(mass times velocity).The second law of motion itself states that force is the rate of change of momentum.How can you know about force but not about momentum?
jfizzix
#6
Sep13-13, 05:01 PM
P: 221
A lot of the time the net force on an object is taught as equal to its mass times the derivative of its velocity instead of derivative of mass times velocity, especially at freshman level physics.

Momentum is more than just total of M*V. It is, "The thing which is conserved because of Newton's third law". Light has momentum, but we wouldn't describe its momentum as M times V.
quawa99
#7
Sep14-13, 09:26 AM
P: 63
Quote Quote by jfizzix View Post
A lot of the time the net force on an object is taught as equal to its mass times the derivative of its velocity instead of derivative of mass times velocity, especially at freshman level physics.

Momentum is more than just total of M*V. It is, "The thing which is conserved because of Newton's third law". Light has momentum, but we wouldn't describe its momentum as M times V.
Yeah momentum is M*V/√(1-V^2/c^2) .So you expect a freshman level physics student to understand that and about how photon has momentum but can't expect him to know that force is the rate of change of momentum.ok
jfizzix
#8
Sep14-13, 11:49 PM
P: 221
Quote Quote by quawa99 View Post
Yeah momentum is M*V/√(1-V^2/c^2) .So you expect a freshman level physics student to understand that and about how photon has momentum but can't expect him to know that force is the rate of change of momentum.ok
Momentum is not just [itex] \frac{M\vec{v}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/itex] either. Certainly not for a (massless) photon in any case, but that's beside the point.

The point is that momentum is a thing that is not defined by its experessions, but by its total value being conserved via Newton's third law. That where the expressions come from.

Also, for goodness sake, it's barely two weeks into the semester. When I was a freshman in physics I couldn't say I understood much of anything only two weeks into the class. Everyone learns it for the first time somewhere.
jfizzix
#9
Sep15-13, 12:09 AM
P: 221
In the freshman physics curriculum, usually momentum conservation is taught (just) after newton's second law, which is why [itex]\vec{F}=m\vec{a}[/itex] is more ubiquitous than [itex]\vec{F}=\frac{d\vec{p}}{dt}[/itex] (among other reasons).


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