# Help with Homework Statement 1b: Elasticity & Pressure

• Vexysery
In summary, the question involves calculating the average force and pressure on a window pane when 1000 hailstones with a mass of 0.5 g and speed of 10ms-1 strike it at a 45° angle. The answer for the elastic collision is 0.0833N for average force and 0.0833Pa for average pressure. In the case of the hailstones striking soft putty and becoming embedded, the change in momentum would be mv for one stone, resulting in a force of 0.0833N and a pressure of 0.0833Pa.

#### Vexysery

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

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.