Help with Homework Statement 1b: Elasticity & Pressure

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Homework Help Overview

The discussion revolves around a physics problem related to elasticity and pressure, specifically focusing on the behavior of hailstones striking a window pane versus soft putty. The original poster seeks clarification on how the change in momentum and resulting force and pressure would differ in these two scenarios.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the change in momentum for elastic versus inelastic collisions, with some questioning the implications of the hailstones embedding in putty. There is also mention of using Newton's second law to analyze the average force.

Discussion Status

Some participants have provided insights into the change in momentum and its implications for the average force and pressure. However, there is no explicit consensus on the approach to take for part 1b, and the discussion remains open with various interpretations being explored.

Contextual Notes

Participants are navigating the complexities of elastic and inelastic collisions, particularly in the context of a homework problem that may impose specific constraints on the analysis.

Vexysery
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I need help on question 1b here,

Homework Statement



1. In a one-minute interval, 1000 hailstones, each of mass 0.5 g and speed 10ms-1, strike a
window pane of area 1m2 at an angle of 45°.
(a) If the collisions are elastic, calculate :
(i) the average force and
(ii) the average pressure on the window pane.

(b) How would your answer di ffer if, in Q1, instead of striking a window pane, the
hailstones strike soft putty and are embedded in it?

Homework Equations



Δρ = (mv) - (-mv) = 2mv ? now it is not elastic = mv?
P = F/A = N/V (mv^2) = N/v ΔT A (mv^2) = N/ΔT (mv)
V = v ΔT A

The Attempt at a Solution



I worked out that the average force is 0.0833N and the average pressure is 0.0833Pa. I don't think the force changes at all, but the change of momentum obviously does as there is no rebound. I'm stuck with this and am not sure how to find out how it would change in ai) and aii)

Any help is very much appreciated! Thank you
 
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Vexysery said:
Δρ = (mv) - (-mv) = 2mv

Yes if collision is perfectly elastic.
 
Anybody help please? :)
 
For av F try to use Newton's 2nd law.
 
You already have the change in momentum of one particle.
 
I need help for question 1b, nothing else.

Thanks anyway!
 
If the stones get embedded in the putty one can assume that all the initial momentum becomes zero.

Hence change in momentum = mv for one stone, as you yourself said.
 

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