Mechanics
Solution Set II
September 26, 2019
Problem 2 Let the speed of the acrobat when he reaches height h be v. Then
v 2 − v02
= −2gh
q
=
v02 − 2gh
=⇒ v
The (vertical component of) momentum of the acrobat (plus monkey) just before the monkey is picked up is
Pi = M v. The momentum just after the monkey is picked up is Pf = (M + m)v 0 where v 0 is their velocity. From
(approximate) conservation of momentum, Pf = Pi which gives
v0 =
M
v
M +m
At the top, their velocity is zero. If they rise together by height h0 , then
0 − v 02
= −2gh0
1 02
=
v
2g
2 2
v0
M
−h
=
M +m
2g
=⇒ h0
Problem 3
The rocket equation applied to this system gives
M
dv
dM
=−
v0 − µM g
dt
dt
Let the initial mass be M0 . Time t0 after which half the mass is used up is
t0 =
M0
2γ
In an infinitesimal duration dt, the change in velocity satisfies
dv = −
dM
v0 − µgdt
M
Integrating upto time t0 gives
1
v(t0 ) = −v0 ln
− µgt0
2
µgM0
= v0 ln 2 −
2γ
Problem 4
Let the instantaneous mass of the freight car be M (t) and its velocity be v(t). The horizontal component of
momentum of the freight car plus mass ∆m of sand about to fall on to it in duration ∆t is P (t) = M (t)v(t). The
1
momentum of the system at instant t + ∆t is P (t + ∆t) = (M (t) + ∆m)v(t + ∆t). The change in momentum of the
system is
∆P
= P (t + ∆t) − P (t)
= M ∆v + ∆M v
Then, the rate of change of momentum is
dP
dt
dv dM
+
v
dt
dt
=
M
=
d
(M v)
dt
Equating this to the external horizontal force gives
d
(M v) = f
dt
Integrating this gives
(M + m)v − 0
=⇒ v
= Ft
Ft
=
M +m
The time it takes for mass m of sand to fall into the freight car is t = m/b. Substituting this gives
v=
Fm
b(M + m)
Problem 5 We can use the rocket equation, with mass being added to the system, as opposed to lost. The
equation is then
dM
dv
=+
urel
M
dt
dt
which gives
dM
dv = urel
M
Integrating
M (t)
v(t) = urel ln
M0
Given that dM/dt = 10 and t = 100s, the increase in mass is ∆M = 1000kg. The mass at instant t is then
M (t) = M0 + ∆M = 3000kg. This gives (using urel = 5m/s)
3
v(t) = 5 ln
m/s
2
Problem 6 Let the instantaneous velocity of the car be V and its instantaneous mass M . Let a man of mass m
jump off with relative speed u. The initial momentum before the man jumps is Pi = M V . The momentum of the
system after the man jumps off is Pf = (M − m)V 0 + m(V 0 − u) = M V 0 − mu, where V 0 is the velocity of the car
after the man jumps off. Momentum conservation gives
V0 =V +
Then,the change in velocity is
m
u
M
m
u
M
which is independent of the initial velocity. Everytime a man jumps off, this is the velocity change (M is different
for each jump). The initial mass of the car is M0 + N m where N is the total number of men on the car and M0 is
∆V =
2
the mass of the empty car. The velocity changes after successive jumps will then be
m
∆V1 =
u
M0 + N m
m
∆V2 =
u
M0 + (N − 1)m
m
∆V3 =
u
M0 + (N − 2)m
...
m
∆VN =
u
M0 + m
The total change in velocity is then
∆V
=
∆V1 + ∆V2 + ... + ∆VN
1
1
1
1
+
+
+ ... +
= mu
M0 + N m M0 + (N − 1)m M0 + (N − 2)m
M0 + m
If all the men jump off at once, the change in velocity is
Nm
u
∆V 0 =
M0 + N m
1
1
1
1
= mu
+
+
+ ... +
M0 + N m M0 + N m M0 + N m
M0 + N m
Clearly, ∆V > ∆V 0 .
Problem 7 In time ∆t, the amount of water shot out is ∆m = K∆t. The speed of water when it reaches height h is
q
v = v02 − 2gh
In time ∆t, the same amount of water hits the can as is ejected out (assuming water to be incompressible). The
maximum change in momentum is imrted to the can when this mass elastically bounces off. This change in
momentum is
∆Pmax = 2∆mv
Then, the force maximum exerted on the can is
Fmax
=
=
=
∆Pmax
∆t
2Kvq
2K
v02 − 2gh
This force balances the weight W of the can. Then, the height of the can satisfies
q
2K v02 − 2gh = W
Problem 8 Let the instantaneous mass of the drop be M (t) and its velocity V (t). Within the time interval ∆t, the
drop accumulated mass ∆M and its velocity changes to V (t + ∆t). The change in the momentum is then
∆P
=
(M + ∆M )V (t + ∆t) − M V (t)
= M ∆V + ∆M V
The rate of change of momentum equals the force acting on the drop due to gravity. Then, the equation of motion
of the drop is
dV
dM
+
V
dt
dt
dV
=⇒ M
+ kM V 2
dt
dV
=⇒
+ kV 2
dt
M
3
= Mg
= Mg
= g
This is a simple first order differential equation which can be integrated
Z
0
V (t)
1
dV
g − kV 2
Z
t
=
0
= t
The integral is simple and gives
1
√ ln
2 gk
1+
p
k/g V
1−
p
k/g V
which can be solved to give
r
V (t) =
√
!
e2 kgt − 1
√
e2 kgt + 1
g
k
√
If t >> 1/ kg, the velocity will be
r
V '
4
g
k
=t
!