relative speed, momentum

a person throws snowballs at a car at a rate R kg/s at a speed u
v is the speed of the car going away from the person

i know that the the relative speed between the car and ball is u-v which means the rate at which it hits the ball is lower but why is the rate R(u-v)/u ?

The rate oif throwing is not the same thing as the speed. The rate will be in kg/time unit whilst speed is in distance/time unit.

i know that, but why is the rate at which it hits the car R(u-v)/u kg/s ?

The distance which the snowball must travel in any given second is $(u-v) \ m$
It's speed is $u \ ms^{-1}$
Does that help?

I think the easiest way to visualise this is just consider the quotient as a ratio. It tells you at speed $u \,\,ms^{-1}$ the rate is $R \,\, kg/s$. The relative velocity is clearly less than u, thus the new rate will be less than R. To find the new rate, we just need the ratio between the relative velocity and u. Hence $R \dfrac{(u-v)}{u}$.

This is also dimensionally correct, as the units are still $kgs^{-1}$


Posted from TSR Mobile

is there a more direct derivation? this just sounds like guessing to me- ' the relative velocity u-v is lower than u so the rate must be lower by the ratio of them, (u-v)/u ' the answer does make sense but how you get to it isn't completely clear

Perhaps the way I worded it seemed like guessing, but nevertheless, I'm pretty sure it makes perfect sense. I was trying to explain it in more layman terms to get the idea across.

If the person is throwing snow balls at rate $R \, kgs^{-1}$ at speed $u \, ms^{-1}$, the rate for a velocity which is less than u will be less than R clearly. It's linear, for example, if $v = \dfrac{u}{2}$, you'd get half the amount of snowballs hitting per second (takes twice as long to hit the car), which can be easily seen as the ratio, $R\dfrac{(\dfrac{u}{2})}{u} \, \, \Rightarrow \, \dfrac{1}{2}R$