Complete Inelastic collision & Force

AI Thread Summary
The discussion revolves around calculating the average force exerted by raindrops on a skylight during a completely inelastic collision. Given the mass of each raindrop and their constant speed, the average force was initially miscalculated but later corrected to 0.00891 N by considering only the momentum transfer of mv, as the drops do not bounce back. The pressure on the window can then be derived from this force and the window's area, leading to a straightforward calculation. The confusion stemmed from the nature of the collision, emphasizing that inelastic collisions result in the raindrops stopping upon impact. Understanding these principles is crucial for solving similar physics problems effectively.
begbeg42
Messages
26
Reaction score
0

Homework Statement



Raindrops of mass 0.99 mg fall vertically at a constant speed of 9.0 m/s, striking a horizontal skylight at the rate of 1000 drops/s and draining off. The window size is 17.0 cm x 25.0 cm. Assume the collisions of the drops with the window are completely inelastic.

(a) Calculate the magnitude of the average force of the raindrops on the window.
(b) Calculate the resulting pressure developed by the raindrop collisions.

Homework Equations


P=F/A
F=dp/dt=Impulse/(delta-time)
Impulse=delta-p=pf-pi


The Attempt at a Solution



so we know that every 1/1000 second, 1 raindrop falls and this is the delta-time
I thought since momentum is conserved the impulse would just be twice mv=2(9.9E-7 kg)(9 m/s)=1.782E-5 Ns
then the Faverage per raindrop= 1.782E-5/(1/1000)=.01782 N...but this is not right

and I know part b is dependent on (a) since then the pressure is simply Faverage/Area= Faverage/(.0425 m^2), where .0425 m^2=area of window=.17 m x .25 m

I'm not sure what to make of the fact that it's completely inelastic and it drains off?
thank you in advance
 
Physics news on Phys.org
The momentum transferred would be 2mv if the drops hit the glass and bounced back with the same speed they came in with, i.e. if the collision was elastic. But in this case they just hit it and stop, so how much momentum is transferred to the glass?
 
just mv? then
Fave=mv/delta-t?=8.91E-6/(1/1000s)=0.00891 N
 
ah thank you! what a stupid question.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top