Download ps4_sol - Physics 141, Fall 2003

Problem 3.18
A raindrop of initial mass M 0 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 its
instantaneous mass and its instantaneous velocity:
dM
 kMV , where k is a constant.
dt
Show that the speed of the drop eventually becomes effectively constant, and give an
expression for the terminal speed. Neglect air resistance.
Solution:
According to equation (3.11),

 dP
,
F
dt




where F  Mg , and P  MV
write the equation in its scalar form, one obtains:
d ( MV )
dV
dM
Mg 
M
V
dt
dt
dt
dM
 kMV , we see that:
since
dt
dV
Mg  M
 kMV 2
dt
dV
g
that is: kV 2 
dt
Then we can see, as the drop falls, it’s velocity increases due to gravity. However, there
is a limit for this velocity because as the drop speeds up, its acceleration decreases rapidly
and will become negligible as the time goes to infinity. So the speed of the drop will
eventually become constant.
dV
0
To compute the terminal constant speed, we could set
dt
Then we have: kVt 2  g
Or
Vt  g
k
Problem 4.2
A block of mass M slides along a horizontal table with speed v0 . At x  0 it hits a spring
with spring constant k and begins to experience a friction force. The coefficient of
friction is a variable and is given by   bx , where b is a constant. Find the loss in
mechanical energy when the block has first come momentarily to rest.
Solution:
Write the position where the block has first come momentarily to rest as x1 . So v1  0
According to the work-energy theorem:
x1
1
1
1
1
2
2
2
2
Mv1  Mv0  0  Mv0   Mv0   F ( x)dx (*)
2
2
2
2
0
and we know that F ( x)  kx  uMg  kx  bMgx
then (*) is just:
x1
1
1
2
 Mv0  (k  bMg )  xdx   (k  bMg ) x12
2
2
0
or
x1  M
v
(k  bMg ) 0
the mechanical energy we have at this point:
kMv02
1
E '  K  U  0  kx12 
2
2(k  bMg )
the mechanical energy we lost:
kMv02
1
bM 2 g
2
E  E '  Mv0 

v02
2
2(k  bMg ) 2(k  bMg )
Problem 4.3
A simple way to measure the speed of a bullet is with a ballistic pendulum. This consists
of a wooden block of mass M into which the bullet is shot. The block is suspended from
cables of length l, and the impact of the bullet causes it to swing through a maximum
angle  . 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 that
this happens quickly)
b. Show how to find the velocity of the bullet by measuring m, M, l, and  .
Solution:
a. As we assume that this happens quickly, total momentum is conserved. That is:
mv  ( M  m)v' , where v ' is the speed of the block and the bullet after the “collision”.
Solve this equation, we get:
mv
v' 
M m
b. As the block swings, its mechanical energy is conserved. So we have:
1
E  Mv '2  Mgl  0  Mgl cos 
2
or
v'  2 gl (1  cos  )
plug in the equation for v ' obtained in part a, we have:
mv
 2 gl (1  cos  )
M m
or
M m
v
2 gl (1  cos  )
m
Problem 4.5
Mass m whirls on a frictionless table, held to circular motion by a string which passes
through a hole in the table. The string is slowly pulled through the hole so that the radius
of the circle changes from l1 to l 2 . Show that the work done in pulling the string equals
the increase in kinetic energy of the mass.
Solution:
Suppose that the mass has velocity v1 when the radius is l1 , v2 when the radius is l 2 ,
vr when the radius is r.



Newton’s 2nd law: F  ma , where a  ar rˆ  aˆ as in plane polar coordinate system.

Here as we are pulling the string slowly, we have F   Fr rˆ , as Fr is the amplitude of the
force. (note: the force is pointing to the origin, which gives us a minus sign as here we
choose r̂ to be our positive direction)
So we know that: Fr  mar , and a  0 .

As in plane polar coordinate system, a  (r  r 2 )rˆ  (r  2r)ˆ
1 d (r 2)
We get that Fr  m(r  r2 ) , and r  2r 
0
r dt
From the 2nd equation, we know that r 2 , which is just rvr , is a constant.
vl
vl
So vr  1 1 , v2  1 1
r
l2
As we pull the string really slow, we could take r  0 . Then we have:
v2
v 2l 2
Fr  mr 2  m r  m 1 31
r
r
so the work we have done in pulling the string:
l2
l2
 
1
v 2l 2
1
1
W   F  dr    m 1 31 dr  mv12l12 ( 2  2 ) , as we know  r  3   r  2
2
r
2l2 2l1
l1
l1
and the increase in kinetic energy of the mass:
1
1
1 v 2l 2
1
1
E  mv22  mv12  m( 1 21  v12 )  mv12l12 ( 2  2 )  W
2
2
2
l2
2l2 2l1
Problem 4.12
During the Second World War the Russian, lacking sufficient parachutes for airborne
operations, occasionally dropped soldiers inside bales of hay onto snow. The human body
can survive an average pressure on impact of 30 lb / in 2 .
Suppose that the lead plane drops a dummy bale equal in weight to a loaded one from an
altitude of 150 ft, and that the pilot observes that it sinks about 2 ft into the snow. If the
weight of an average soldier is 144 lb and his effective area is 5 ft 2 , is it safe to drop the
men?
Solution:
Suppose we drop the dummy bale at ha , it stops at last at hb . And for the ground we have
h  0.
the work-energy theorem gives
b 

Kb  K a   F  dR
a

if we think of an average N as the force the snow acts on the dummy bale after it hits the
ground, the equation is just:
0  0  mg(hb  ha )  N (hb  0)
mg (hb  ha )
that is: N 
hb
N mg (hb  ha )
the pressure will be: P  
S
Shb
as m  144lb, g  9.8m / s 2 , hb  2 ft, ha  150 ft, S  5 ft 2
144  (2  150)
we get P 
lb / ft 2  g  15.2lb / in 2  g , where we have used the fact
5  (2)
1
ft

12
in
that
as the human body can survive an average pressure on impact of 30 lb / in 2 , which is
almost twice as big as the pressure on the dummy bale, we could say it is safe to drop the
men.