Problem 3.18A raindrop of initial mass 0M starts falling from rest under the influence of gravity.

Assume that the drop gains mass from the cloud at a rate proportional to the product of itsinstantaneous mass and its instantaneous velocity:

€

dMdt

= kMV , where k is a constant.

Show that the speed of the drop eventually becomes effectively constant, and give anexpression for the terminal speed. Neglect air resistance.

Solution:According to equation (3.11),

dt

PdF

rr= ,

where gMFrr

= , and VMPrr

=write the equation in its scalar form, one obtains:

dt

dMV

dt

dVM

dt

MVdMg +==

)(

since kMVdt

dM= , we see that:

2kMVdt

dVMMg +=

that is: gdt

dVkV =+2

Then we can see, as the drop falls, it’s velocity increases due to gravity. However, thereis a limit for this velocity because as the drop speeds up, its acceleration decreases rapidlyand will become negligible as the time goes to infinity. So the speed of the drop willeventually become constant.

To compute the terminal constant speed, we could set 0=dt

dV

Then we have: gkVt =2

Or

kgVt =

Problem 4.2

A block of mass M slides along a horizontal table with speed 0v . At 0=x it hits a spring

with spring constant k and begins to experience a friction force. The coefficient offriction is a variable and is given by bx=µ , where b is a constant. Find the loss inmechanical energy when the block has first come momentarily to rest.

Solution:Write the position where the block has first come momentarily to rest as 1x . So 01 =vAccording to the work-energy theorem:

∫=−=−=−1

0

20

20

20

21 )(

2

1

2

10

2

1

2

1x

dxxFMvMvMvMv (*)

and we know that bMgxkxuMgkxxF −−=−−=)(then (*) is just:

21

0

20 )(

2

1)(

2

1 1

xbMgkxdxbMgkMvx

+−=+−=− ∫or

01 )( vbMgkMx +=

the mechanical energy we have at this point:

)(22

10

202

1'

bMgk

kMvkxUKE

+=+=+=

the mechanical energy we lost:

20

2202

0 )(2)(22

1' v

bMgk

gbM

bMgk

kMvMvEE

+=

+−=−

Problem 4.3

A simple way to measure the speed of a bullet is with a ballistic pendulum. This consistsof a wooden block of mass M into which the bullet is shot. The block is suspended fromcables of length l, and the impact of the bullet causes it to swing through a maximumangle φ . The initial speed of the bullet is v, and its mass is m.

a. How fast is the block moving immediately after the bullet comes to rest? (assume thatthis happens quickly)

b. Show how to find the velocity of the bullet by measuring m, M, l, and φ .

Solution:a. As we assume that this happens quickly, total momentum is conserved. That is:

')( vmMmv += , where 'v is the speed of the block and the bullet after the “collision”.Solve this equation, we get:

mM

mvv

+='

b. As the block swings, its mechanical energy is conserved. So we have:

φcos0'21 2 MglMglMvE −=−=

or)cos1(2' φ−= glv

plug in the equation for 'v obtained in part a, we have:

)cos1(2 φ−=+

glmM

mv

or

)cos1(2 φ−+

= glm

mMv

Problem 4.5

Mass m whirls on a frictionless table, held to circular motion by a string which passesthrough a hole in the table. The string is slowly pulled through the hole so that the radiusof the circle changes from 1l to 2l . Show that the work done in pulling the string equalsthe increase in kinetic energy of the mass.

Solution:Suppose that the mass has velocity 1v when the radius is 1l , 2v when the radius is 2l ,

rv when the radius is r.

Newton’s 2nd law: amFrr

= , where θθ ˆˆ araa r +=r

as in plane polar coordinate system.

Here as we are pulling the string slowly, we have rFF r ˆ−=r

, as rF is the amplitude of theforce. (note: the force is pointing to the origin, which gives us a minus sign as here wechoose r̂ to be our positive direction)So we know that: rr maF = , and 0=θa .

As in plane polar coordinate system, θθθθ ˆ)2(ˆ)( 2 &&&&&&&r

rrrrra ++−=

We get that )( 2θ&&& rrmFr −−= , and 0)(1

22

==+dtrd

rrr

θθθ

&&&&&

From the 2nd equation, we know that θ&2r , which is just rrv , is a constant.

So rlv

vr11= ,

2

112 l

lvv =

As we pull the string really slow, we could take 0=r&& . Then we have:

3

21

21

22

rlv

mrv

mmrF rr === θ&

so the work we have done in pulling the string:

)21

21

( 21

22

21

213

21

21

2

1

2

1ll

lmvdrrlv

mrdFWl

l

l

l

−=−=⋅= ∫∫rr

, as we know ∫ −− −= 23

21rr

and the increase in kinetic energy of the mass:

Wll

lmvvllv

mmvmvE =−=−=−=Δ )21

21

()(21

21

21

21

22

21

21

212

2

21

212

122

Problem 4.12

During the Second World War the Russian, lacking sufficient parachutes for airborneoperations, occasionally dropped soldiers inside bales of hay onto snow. The human bodycan survive an average pressure on impact of 30 2/ inlb .

Suppose that the lead plane drops a dummy bale equal in weight to a loaded one from analtitude of 150 ft, and that the pilot observes that it sinks about 2 ft into the snow. If theweight of an average soldier is 144 lb and his effective area is 5 2ft , is it safe to drop themen?

Solution:Suppose we drop the dummy bale at ah , it stops at last at bh . And for the ground we have

0=h .the work-energy theorem gives

∫ ⋅=−b

a

ab RdFKKrr

if we think of an average Nr

as the force the snow acts on the dummy bale after it hits theground, the equation is just:

)0()(00 −+−−=− bab hNhhmg

that is: b

ab

hhhmg

N)( −

=

the pressure will be: b

ab

Shhhmg

SN

P)( −

==

as 22 5,150,2,/8.9,144 ftSfthfthsmglbm ab ==−===

we get ginlbgftlbP ×=×−×−−×

= 22 /2.15/)2(5

)1502(144, where we have used the fact

that inft 121 =

as the human body can survive an average pressure on impact of 30 2/ inlb , which isalmost twice as big as the pressure on the dummy bale, we could say it is safe to drop themen.