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Chapter 6: Q50P (page 1356)

A particle is confined within a box with perfectly rigid walls at x= 0 and x= L. Although the magnitude of the instantaneous force exerted on the particle by the walls is infinite and the time over which it acts is zero, the impulse (that involves a product of force and time) is both finite and quantized. Show that

the impulse exerted by the wall at x= 0 is (nh/L)i^ and that the impulse exerted by the wall at x= Lis - (nh/L)i^. (Hint:You may wish to review Section 8.1.)

Short Answer

Expert verified

It is shown that the impulse exerted by the wall at x= 0 is nh/Li^ and that the impulse exerted by the wall at x= Lis -nh/Li^.

Step by step solution

01

(a) Identification of the concept.

According to Newton鈥檚 laws of motion,

The impulse acting on a surface by a particle is equal to the change of its momentum.

02

(b) Determination of the impulse exerted by the wall at x = 0 and x = L.

The magnitude of the discrete momentum of a particle in box is,

p=魔办=nh2L

The change in momentum is,

鈭�p鈫�=p鈫�final-p鈫�initialp鈫�=魔苍罢罢L=hn2L

Now, the initial momentum at the wall at x = 0 is,

p鈫�initial=-hn2Li铃�

And the final momentum at the wall at x = 0 is,

p鈫�final=-hn2Li铃�

So, the change in momentum is,

localid="1664001916349" 鈭�p=+hn2Li铃�--hn2Li铃�=+hn2Li铃�

Similarly, the initial momentum at the wall at x = L is,

p鈫�initial=+hn2Li铃�

And the final momentum at the wall at x = L is,

p鈫�final=-hn2Li铃�

So, the change in momentum is,

鈭�p=-hn2Li铃�-hn2Li铃�=-hnLi铃�

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