Help with Homework Statement 1b: Elasticity & Pressure

Click For Summary
SUMMARY

The discussion focuses on Homework Statement 1b regarding the physics of hailstones striking a window pane and then soft putty. The average force calculated for the elastic collision scenario is 0.0833 N, with an average pressure of 0.0833 Pa on the window pane. In the case of the hailstones embedding in putty, the change in momentum is simply mv, as the initial momentum becomes zero upon impact. The participants emphasize the importance of understanding elastic versus inelastic collisions in calculating these values.

PREREQUISITES
  • Understanding of elastic and inelastic collisions
  • Knowledge of Newton's 2nd Law
  • Familiarity with momentum calculations
  • Basic concepts of pressure and force
NEXT STEPS
  • Study the principles of elastic and inelastic collisions in depth
  • Learn how to apply Newton's 2nd Law to various scenarios
  • Explore momentum conservation in different collision types
  • Investigate pressure calculations in fluid dynamics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and collision theory, as well as educators looking for examples of elastic and inelastic collisions in real-world applications.

Vexysery
Messages
4
Reaction score
0
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
 
Physics news on Phys.org
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.
 

Similar threads

Replies
19
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 13 ·
Replies
13
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K