Page History

Part

B

Consider

again

the

situation

of

Part

A.

Suppose

that

the

rain

is

falling

with

a

speed

of

10.0

m/s,

and

that

it

comes

to

rest

with

respect

to

the

train

car

immediately

upon

impact.

At

the

instant

the

rain

begins

to

hit

the

car,

what

is

the

force

exerted

on

the

train

car

by

the

rain?

h4. Solution *System:* The rain entering the car in a time _t_, treated as a [point particle]. *Interactions:* External forces from gravity and from the railroad car. We assume the collision force is much larger than the force of gravity during the collision. *Model:* [Momentum and External Force]. *Approach:* {note} Note that we have chosen to model the water, even though we are asked for the force from the water on the car. During the collision, the collision forces on the water dominate and so we can solve for them. The collision forces on the car, however, are not dominant. The y-component of the collision force is completely balanced by the normal force. For this reason it is hard to solve for the force using the car's dynamics. Instead, we will use [Newton's 3rd Law|Newton's Third Law] to relate the collision force on the rain to the force from the rain on the car.{note} We will make the approximation that enough raindrops are hitting the car that the rain can essentially be treated as a continuous flow of water. The force that acts on the water is equal to the time derivative of its momentum. Because of the discontinuous nature of the impact (the rain is assumed to stop instantaneously) it is impossible to express the rain's momentum as a continuous function of time. For this reason, we must re-express the law of interaction: {latex}\begin{large}\[ F_{x} = \frac{dp_{x}}{dt} = \lim_{t \rightarrow 0} \frac{p_{x}(t) - p_{x}(0)}{t} = \lim_{t \rightarrow 0} \frac{m^{\rm rain}(t)v^{\rm car} - m^{\rm rain}(t)(\mbox{0 m/s})}{t}\] \[F_{y} = \frac{dp_{y}}{dt} = \lim_{t \rightarrow 0} \frac{p_{y}(t) - p_{x}(0)}{t} = \lim_{t \rightarrow 0} \frac{m^{\rm rain}(t)(\mbox{0 m/s}) + m^{\rm rain}(t)v^{\rm rain}}{t}\]\end{large}{latex} {note}Note that the rain is falling downward, which we have chosen to be the negative _y_ direction.{note} The mass of rain that enters the car in a time _t_ can be expressed in terms of the rate at which the rain accumulates in the car, known to be: {latex}\begin{large}\[ r = \frac{\mbox{1.0 cm}}{\mbox{1.0 min}} = \frac{\mbox{0.01 m}}{\mbox{60 s}} = \mbox{0.00017 m/s} \]\end{large}{latex} {warning}Because rain falls in drops, not as a continuous sheet, the rate of height accumulation is *not* the same as the rain's speed through the air!{warning} In terms of this rate _r_, we can write: {latex}\begin{large}\[ m^{\rm rain}(t) = \rho^{\rm water}rlwt \]\end{large}{latex} where _l_ and _w_ are the length and width of the train car. We now have: {latex}\begin{large}\[ F_{x} = \lim_{t \rightarrow 0} \frac{\rho^{\rm water}rlwtv^{\rm car}}{t} = \rho^{\rm water}rlwv^{\rm car} = 12.5 N \] \[ F_{y} = \lim_{t \rightarrow 0} \frac{\rho^{\rm water}rlwtv^{\rm rain}}{t} = \rho^{\rm water}rlwv^{\rm rain} = 50 N \]\end{large}{latex} {tip}Both components should be positive, since the rain is _acquiring a positive_ x-velocity and _losing a negative_ y-velocity.{tip} The total force on the rain is therefore: {latex}\begin{large} \[ F^{\rm rain} = \mbox{12.5 N}\hat{x} + \mbox{50 N}\hat{y} = \mbox{52 N at 76}^{\circ}\mbox{ counterclockwise from the +x axis}\]\end{large}{latex} Thus, the force on the _car_ is 52 N at 256° counterclockwise from the +x axis.