Momentum/Impulse homework question

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Homework Help Overview

The problem involves a scenario where a child on roller blades pushes against a wall, resulting in motion affected by friction. The subject area encompasses concepts of momentum, impulse, and forces, particularly in the context of dynamics and kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations related to the child's motion after pushing off the wall, questioning the distinction between forces during the push and the subsequent motion influenced by friction. There are attempts to clarify the relationship between impulse and momentum, with some participants expressing uncertainty about the implications of their calculations.

Discussion Status

The discussion is ongoing, with participants providing insights and corrections regarding the original poster's calculations. There is a focus on understanding the impulse generated during the push and how it relates to the child's change in momentum. Multiple interpretations of the problem are being explored, particularly concerning the timing of forces and the role of friction.

Contextual Notes

Participants are navigating the complexities of the problem, including the initial conditions of motion and the effects of friction. There is an acknowledgment of potential misunderstandings regarding the definitions of impulse and momentum, as well as the timing of events in the scenario.

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Homework Statement



A 46.5 kg kid on roller blades pushes as hard as he can on a wall which causes him to start rolling away from the wall. At this point, friction begins to slow him down and he comes to rest 14.3 m from the wall. The coefficient of friction is 0.125. If his push lasted 0.460 s, how hard did he push on the wall (in Newtons)?

Rough diagram:
PAxhj.png


2. The attempt at a solution

J = Δ p
fΔt = mv2 - mv1 (mv2 = 0, since v2 is 0)
v1 = (Fft)/(-46.5)
v1 = -0.563 m/s **

Ff = μFn
=(0.125)(46.5)(9.8)
= 56.96N ( **I substituted this value to Ff in the (Fft)/(-46.5) equation to find v1)

a = v2 - v1 (v2 = 0)
t​
a =0.563
0.460​
a= 1.225 m/s2

Fnet = ma
= (46.5)(1.225)
= 56.96N
Therefore, the average force exerted on the wall is 56.96N(BKWD)?

I tried doing this problem today but I feel that the answer is wrong(doesn't make sense) and that I made a mistake. Any help would greatly be appreciated :-p
 
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It looks like all your calculations are for the period during which the kid is rolling. I.e., the acceleration you found is due to friction, not the force of the wall (in fact, at this point he's already let go of the wall)
 
JaWiB said:
It looks like all your calculations are for the period during which the kid is rolling. I.e., the acceleration you found is due to friction, not the force of the wall (in fact, at this point he's already let go of the wall)

So what do I need to change/add?

I understand what you are telling me, but I don't know what do I do
 
His initial momentum is zero (I guess), when he's done pushing on the wall, he has the velocity that you calculated. So what's the impulse?
 
JaWiB said:
His initial momentum is zero (I guess), when he's done pushing on the wall, he has the velocity that you calculated. So what's the impulse?

Impulse = Δ p
and since Δ p = 0, then J = Ff x t which is 56.96 x 0.460 which gives you 26.20 N.s

What do I do with the impulse value that I got?
 
Last edited:
Bump, can anyone help me out here?
 
Why did you say the change in momentum is zero? Before he pushes off the wall, he has no speed, and after he has some nonzero speed and therefore nonzero momentum. And if the change in momentum were zero, then J = 0.

J = F_push * t_push

You have t_push, you can calculate J from the change in momentum (which you can find using the velocity you calculated) so then you just have to solve for F_push
 

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