Question

Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of 15 m/s, and the mass of rain per second striking the roof is 0.071 kg/s. (a) Assuming the drops come to rest after striking the roof, find the average force exerted by the rain on the roof. (b) If hailstones having the same mass as the raindrops fall on the roof at the same rate and with the same speed, how would the average force on the roof compare to that found in part (a)?

Answer 1

Answer: (a) 1.065 N (b) 2.13 N

Step-by-step explanation:

(a) average force exerted by the rain on the roof

According Newton's 2nd Law of Motion the force
`F` is defined as the variation of linear momentum
`p` in time:

`F=(dp)/(dt)` (1)

Where the linear momentum is:

`p=mV` (2) Being
`m` the mass and
`V` the velocity.

In the case of the rain drops, which initial velocity is
`V_(i)=15m/s` and final velocity is
`V_(f)=0` (we are told the drops come to rest after striking the roof). The momentum of the drops
`p_(drops)` is:

`p_(drops)=mV_(i)+mV_(f)` (3)

If
`V_(f)=0`, then:

`p_(drops)=mV_(i)` (4)

Now the force
`F_(drops)` exerted by the drops is:

`F_(drops)=(dp_(drops))/(dt)=(d)/(dt)mV_(i)` (5)

`F_(drops)=(dm)/(dt)V_(i)+m(dV_(i))/(dt)` (6)

At this point we know the mass of rain per second (mass rate)
`(dm)/(dt)=0.071 kg/s` and we also know the initial velocity does not change with time, because that is the velocity at that exact moment (instantaneous velocity). Therefore is a constant, and the derivation of a constant is zero.

This means (6) must be rewritten as:

`F_(drops)=(dm)/(dt)V_(i)` (7)

`F_(drops)=(0.071 kg/s)(15m/s)` (8)

`F_(drops)=1.065kg.m/s^(2)=1.065N` (9) This is the force exerted by the rain drops on the roof of the car.

(b) average force exerted by hailstones on the roof

Now let's assume that instead of rain drops, hailstones fall on the roof of the car, and let's also assume these hailstones bounce back up off after striking the roof (this means they do not come to rest as the rain drops).

In addition, we know the hailstones fall with the same velocity as the rain drops and have the same mass rate.

So, in this case the linear momentum
`p_(hailstones)` is:

`p_(hailstones)=mV_(i)+mV_(f)` (9) Being
`V_(i)=V_(f)`

`p_(hailstones)=mV+mV=2mV` (10)

Deriving with respect to time to find the force
`F_(hailstones)` exerted by the hailstones:

`F_(hailstones)=(d)/(dt)p_(hailstones)=(d)/(dt)(2mV)` (10)

`F_(hailstones)=2(d)/(dt)(mV)=2((dm)/(dt)V+m(dV)/(dt))` (11)

Assuming
`(dV)/(dt)=0`:

`F_(hailstones)=2((dm)/(dt)V)` (12)

`F_(hailstones)=2(0.071 kg/s)(15m/s)` (13)

Finally:

`F_(hailstones)=2.13kg.m/s^(2)=2.13N` (14) This is the force exerted by the hailstones

Comparing (9) and (14) we can conclude the force exerted by the hailstones is two times greater than the force exerted by the raindrops.