# Chapter 18: Impulse and Momentum¶

## Example 18.18-8, Page No 435¶

In [1]:
import math

#Initilization of variables
W=100 #lb
u=0.2 #coefficient of friction
t=5 #s
v1=5 #ft/s
v2=10 #ft/s
g=32.2 #ft/s**2
ll=0 #lower limit of integration
ul=5 #upper limit of integration

#Calculations
Fr=u*W #lb
#Using The impulse momentum theorem
#Since the integration is just subtraction of limits we can skip that
F=((W/g)*v2-(W/g)*v1+Fr*ul)/ul #lb

#Result
print'The Force is',round(F,1),"lb"

The Force is 23.1 lb


## Example 18.18-9, Page No 435¶

In [15]:
import math

#Initilization of variables
m=10 #kg
#theta=30,thus
cos30=sqrt(3)*2**-1
sin30=2**-1
u=0.3 #coefficient of kinetic friction
t=5 #s
g=9.8 #m/s**2

#Calculations
#Asthere is no motion in the vertical direction
#Summing forces along vertical direction
Na=m*g*cos30 #N
#Using impulse momentum theorem
vx=(m*g*sin30-u*Na)*(t*m**-1) #m/s

#Result
print'The speed after 5s is',round(vx,1),"m/s"

The speed after 5s is 11.8 m/s


## Example 18.18-10, Page No 436¶

In [5]:
import math
def integrand(t, a, b):
return 20*t-16
a=1
b=1
g=32.2 # ft/s**2
v=(F[0]*32.2)/80

# Results
print'The speed of the block is',round(v,1),"ft/s"

The speed of the block is 70.8 ft/s


## Example 18.18-12, Page No 436¶

In [1]:
import math

#Initilization of variables
m1=40 #kg
m2=10 #kg
m3=15 #kg
v0=2.5 #m/s
vf=5 #m/s
t=12 #s
u=0.1 #coefficient of friction
g=9.8 #m/s**2
#theta=45 degrees,thus
sin45=(sqrt(2))**-1
cos45=(sqrt(2))**-1

#Calculations
#Applying Impulse-Momentum Theoroem
P=(((m1+m2+m3)*(vf-v0))+(t*(-m2*g*sin45+u*g*m2*cos45+u*g*m3+g*m1)))*t**-1 #N

#Result
print'The value of P is',round(P),"N"

The value of P is 358.0 N


## Example 18.18-13, Page Noo 437¶

In [24]:
import math

#Initilization of variables
W1=4 #lb
W2=2 #lb
t2=0.04 #s
W3=-2 #lb
t3=0.02 #s
t=3 #s
g=32.2 #ft/s**2

#Calculations
#Algebraic sum of two areas
A=t2*W2+t3*W3 #lb-s
#Using Impulse Momentum Theorem
v=(A*g)/W1 #ft/s

#Result
print'The spped after 3s is',round(v,3),"ft/s"

The spped after 3s is 0.322 ft/s


## Example 18.18-14, Page No 437¶

In [1]:
import math

#Initilization of variables
f_r=1 #in/s rate of fall of mercury
ll=18 #in length of left column
lr=22 #in length of right column
rho=850 #lb/ft**3
d=4**-1 #in
g=32.2 #ft/s**2

#Calculations
#Applying Impulse momentum theorem
M=((d*pi*d**2*4)/12**3)*(rho/g)*(12**-1) #lb-s

#Result
print'The upward momentum of mercury is +',round(M,5),"lb-s"

The upward momentum of mercury is + 0.00025 lb-s


## Example 18.18-15, Page No 437¶

In [27]:
import math

#Initilization of variables
m=2000 #kg
k=1.200 #m
w=120 #rpm
t=200 #s

#Calculations
#Applying Angular Momentum theorem
M=((m*k**2*(w*2*pi))/60)/t #N.m

#Result
print'The Momentum necessary is',round(M),"N.m"

The Momentum necessary is 181.0 N.m


## Example 18.18-17, Page No 438¶

In [2]:
import math

#Initilization of variables
m1=5 #kg
m2=7 #kg
mp=5 #kg
r=0.6 #m
k=0.45 #m
vi=3 #m/s
vf=6 #m/s
g=9.8 #m/s**2

#Calculations
I=m1*k**2 #kg.m**2
#Solving the system of linear equations
#Simplfying the equation we get
t=((I*wnet)+m1*(vf-vi)+m2*(vf-vi))*r/(r*(m2-m1)*g) #s

#Result
print'The time required is',round(t,2),"s"
# The ans in the textbook is incorrect.

The time required is 2.1 s


## Example 18.18-18, Page No 438¶

In [6]:
import math

#Initilization of variables
m=50 #kg
vo=4 #m/s
vf=8 #m/s
t=6 #s
g=9.8 #m/s**2
r=0.8 #m
u=0.25 #coefficient of friction
I=30 #kg-m**2

#Calculations
Na=m*g #N
F=u*Na #N
#Angular Speeds
#Applying impulse momentum theorem
mb=(I*wo+m*vo*r-F*r*t-I*wf-m*vf*r)/(vf*r-vo*r-g*r*t) #kg

#Result
print'The mass of block B is',round(mb,1),"kg"

The mass of block B is 20.5 kg


## Example 18.18-19, Page No 439¶

In [14]:
import math

#Initilization of variables
Ws=250 #lb
Wl=500 #lb
W3=161 #lb
W4=64.4 #lb
wo=100 #rpm
wf=300 #rpm
rl=3 #ft
rs=2 #ft
g=32.2 #ft/s**2

#Calculations
#Moment Of Inertia
I=(0.5*(Wl/g)*rl**2+0.5*(Ws/g)*rs**2) #slug-ft**2
#Change in angular Momentum
change1=I*((wf-wo)*2*(pi/60)) #lb-s-ft
#Change in angular momentum about G for 161lb
change2=2*((W3*g**-1)*(wf-wo)*(4*60**-1)*pi) #lb-s-ft
#Similarly change in 64lb is
change3=3*((W4*g**-1)*(wf-wo)*(6*60**-1)*pi) #lb-s-ft
#Change in linear impulse
#Without t term in it
m1=2*W3
m2=-3*W4
#Total angular impulse
t=(change1+change2+change3)/(m1+m2) #s

#Result
print'The time required is',round(t),"s"

The time required is 20.0 s


## Example 18.18-21, Page No 440¶

In [4]:
import math
import numpy as np

#Initilization of variables
d=3 #ft
W=300 #lb
# as theta=20 degrees
sintheta=0.342
F=250 #lb
t=6 #s
vo=0 #ft/s
g=32.2 #ft/s**2

#Calculations
#Applying linear impulse momentum theorem
#Solving by matrix method
A=np.array([[-W*g**-1,1*t],[-((0.5)*(W*g**-1)*(d*2**-1)**2*(d*2**-1)**-1),(-t)*d*2**-1]])
B=np.array([[-F*t+W*sintheta*t],[-F*(d*2**-1)*6]])
C=np.linalg.solve(A,B)

#Result
print'The speed after 6s is',round(C[0]),"ft/s,parallel to the plane"

The speed after 6s is 171.0 ft/s,parallel to the plane


## Example 18.18-22, Page No 440¶

In [8]:
#Initilization of variables
#theta=30 degrees
sin30=2**-1
vo=20 #ft/s
r=4 #ft
vf=0 #ft/s
W=300 #lb
g=32.2 #ft/s**2

#Calculations
#Applying impulse momentum theorem
#Solving simultaneous equations
t=-((W*g**-1)*(vf-vo)+((0.5*(W*g**-1)*r**2*(wf-wo))*r**-1))/(W*sin30) #s

#Result
print'The time t is',round(t,2),"s"

The time t is 1.86 s


## Example 18.18-23, Page No 441¶

In [59]:
import math
import numpy as np

#Initilization of variables
mw=75 #kg
k=0.9 #m
r=1.2 #m
m=30 #kg
g=9.8 #m/s**2

#Calculations
#Initial speed
vi=-r*wi #m/s
vf=r*wf #m/s
#Initial speed of B is
vib=-0.8*wi+vi #m/s
#Similarly
vfb=12 #m/s
#Applying impulse momentum theorem
#Solving by matrix method
A=np.array([[1,-1,-(mw*(vf-vi))],[0.8,1.2,-(mw*(k**2)*(wf+wi))],[-1,0,-(m*(vfb-vi))]])
B=np.array([[0],[0],[-g*m]])
C=np.linalg.solve(A,B)
#Here t is calculated as 1/t for simplicity

#Result
print'The time required is',round(C[2]**-1,2),"s"
#Decimal accuracy causes discrepancy in answers

The time required is 7.04 s


## Example 18.18-24, Page No 441¶

In [4]:
import math

#Initilization of variables
d=8 #in
W=96.6 #lb
u=0.15 #coefficient of friction
g=32.2 #ft/s**2

#Calculations
r=(d/2)*12**-1 #m
N=W #lb
F=u*N #lb
m=W/g #slugs
I=0.5*m*(r**2) #slug-ft**2
#Applying the impulse momentum theorem
#Solving the two equations simultaneously
A=np.array([[F,-m],[F*r,I*(1*r**-1)]])
B=np.array([[0],[w*I]])
C=np.linalg.solve(A,B)
#Distance travelled
s=0.5*C[1]*C[0] #ft
t=C[0] #s

#Result
print'The time required is',round(t,2),"s",'and it travels',round(s,2),"ft"

The time required is 0.83 s and it travels 1.66 ft


## Example 18.18-25, Page No 442¶

In [9]:
import math

#Initilisation of variables
d=2*12**-1 #ft
v=80 #ft/s
g=32.2 #ft/s**2

#Calculations
#Mass flow reate without time
m=(1*4**-1)*pi*d**2*v*(62.4/g)
#Let P=force of plate on mass m of water
P=m*(0-v) #lb

#Result
print'The force water exerts on the plate is +',round(-P),"lb,that is,to the right"

The force water exerts on the plate is + 271.0 lb,that is,to the right


## Example 18.18-26, Page No 442¶

In [10]:
import math

#Initilization of variables
v1=20 #ft/s
vw=80 #ft/s
d=2*12**-1 #ft
g=32.2 #ft/s**2

#Calculations
v=vw-v1 #ft/s
#mass flow rate without t
m=(4**-1)*(pi*d**2)*(62.4/g)*v
#Applying impulse momentum theorem
P=m*v #lb

#Result
print'The force exerted by water on the plate is +',round(P),"lb,that is,to the right"

# Decimal poinat accuracy causes a small discrepancy in the answer

The force exerted by water on the plate is + 152.0 lb,that is,to the right


## Example 18.18-27, Page No 443¶

In [13]:
import math

#Initilization of variables
A=2000*10**-6 #m**2
v=10 #m/s
rho=1000 #kg/m**3
#theta=45 degrees,thus
cos45=(2**0.5)**-1
sin45=(2**0.5)**-1

#Calculations
#Mass flow
m=A*v*rho
#Applying impulse momentum theorem
Px=m*(-v*cos45-v) #N
Py=m*(v*sin45-0) #N

#Result
print'The x component of force is',round(Px),"N (to left on water)",'and y component is +',round(Py),"N (up on the water)"

The x component of force is -341.0 N (to left on water) and y component is + 141.0 N (up on the water)


## Example 18.18-28, Page No 443¶

In [30]:
import math

#Initilisation of variables
W=150 #lb
v=20 #ft/s
A=0.2 #in**2
t=60 #s
g=32.2 #ft/s**2

#Calculations
#Mass flow
m=(A/12**2)*v*(62.4/g)
#Force
F=m*(0-v) #lb
#At t=60s the tank holds
Ww=(A/12**2)*v*t*62.4 #lb
S=-F+W+Ww #lb

#Result

The scale reading is 255.0 lb


## Example 18.18-29, Page No 444¶

In [32]:
import math

#Initilization of variables
Wp=130 #lb
Wb=150 #lb
Wbullet=2*16**-1 #lb
g=32.2 #ft/s**2
vbullet=1200 #ft/s

#Calculations
v=((-Wbullet*g**-1)*vbullet)/((Wb+Wp)/g) #ft/s

#Result
print'The speed of the boat is',round(v,2),"ft/s"

The speed of the boat is -0.54 ft/s


## Example 18.18-30, Page No 444¶

In [33]:
import math

#Initilization of variables
mb=0.06 #kg
ms=50 #kg
h=0.03 #m
g=9.8 #m/s**2

#Calculations
#Speed of bag+bullet
v2=sqrt(2*g*h) #m/s
#Applying conservation of momentum
v1=((mb+ms)*v2)/mb #m/s

#Result
print'The speed of bullet as it entered the bag was',round(v1),"m/s"

The speed of bullet as it entered the bag was 640.0 m/s


## Example 18.18-32, Page No 445¶

In [34]:
import math

#Initilization of variables
mb=0.06 #kg
vb=500 #m/s
mblock=5 #kg
vblock=30 #m/s

#Calculations
#Applying conservation of momentum
v=(mb*vb+mblock*vblock)/(mb+mblock) #m/s

#Result
print'The speed of the system is',round(v,1),"m/s"

The speed of the system is 35.6 m/s


## Example 18.18-33, Page No 445¶

In [19]:
# Initilization of variables
W1=2 #lb
W2=3 #lb
g=32.2 #ft/s**2

import math
def integrand(x, a, b):
return 12*(2-x)
a=1
b=1
Work=W[0]/12 # ft-lb

# Solving the simultaneousequations
v3=(Work*(0.5*(W2*g**-1)+0.5*(W1*g**-1)*(-W2*W1**-1)**2)**-1)**0.5 #ft/s
v2=-(W2*W1**-1)*v3 #ft/s

# Results
print'The speed of 2lb block is',round(v2,2),"ft/s (to the left)",'and that of 3lb block is',round(v3,2),"ft/s (to the right) respectively."

The speed of 2lb block is -6.22 ft/s (to the left) and that of 3lb block is 4.14 ft/s (to the right) respectively.


## Example 18.18-34, Page No 445¶

In [14]:
import math

#Initilization of variables
#Here the integration is indefinite hence it will be computed manually and entered
W=10 #lb
l=4 #ft
g=32.2 #ft/s**2

#Calculations
#Part (a)
t=(((W/g)*(l*w*l))-((W/g)*(l*wf*l)))**0.5 #s
#Part (b)
#Applying conservation of angular momentum
r=(l*wf*l)*(l*w)**-1 #ft

#Result
print'The answer for part (a) is',round(t,2),"s",'and the answer for part (b) is',round(r,2),"ft"

The answer for part (a) is 1.58 s and the answer for part (b) is 3.0 ft


## Example 18.18-35, Page No 446¶

In [15]:
import math

#Initilization of variables
W=2.5 #lb
Idisk=0.4 #slug-ft**2
g=32.2 #ft/s**2

#Calculations
Ii=Idisk+(2*(W*g**-1)*(3*12**-1)**2) #slug-ft**2
If=Idisk+(2*(W*g**-1)*(11*12**-1)**2) #slug-ft**2
#Since no external moments act
#Applying conservation of momentum

#Result

The final angular speed is 27.8 rad/s


## Example 18.18-39, Page No 448¶

In [52]:
import math

#Initilization of variables
u1=6 #ft/s
u2=-8 #ft/s
e=0.8 #coefficient of restitution

#Calculations
#Solving both simultaneous equations
A=np.array([[1,-1],[1,1]])
B=np.array([[11.2],[-2]])
C=np.linalg.solve(A,B) #ft/s

#Result
print'The velocities are v1=',round(C[1],1),"ft/s",'and v2=',round(C[0],1),"ft/s"

The velocities are v1= -6.6 ft/s and v2= 4.6 ft/s


## Example 18.18-40, Page No 448¶

In [53]:
import math

#Initilization of variables
h1=20 #m
h2=14 #m
g=9.8 #m/s**2

#Calculations
u1=sqrt(2*g*h1) #m/s
u2=0 #m/s
v1=-sqrt(2*g*h2) #m/s
v2=0 #m/s
e=(v2-v1)/(u1-u2) #coefficient of restitution

#Result
print'The value of coefficient of restitution is',round(e,2)

The value of coefficient of restitution is 0.84


## Example 18.18-41, Page No 448¶

In [23]:
import math

#Initilization of variables
u=6.55 #ft/s
g=32.2 #ft/s**2
L=6 #ft
W=5 #lb
c=0.7 #fraction of impulse acting in second phase

#Calculations
#Impulse
I=(W*g**-1)*(u*3**-1) #N.s
#Second Phase
v=((-c*10.9)/2)+u #ft/s

#Result
print'The value is v=',round(v,2),"ft/s to the right",' and that of w is',round(wprime,2),"rad/s counterclockwise."

#The value of w is incorrect in the textbook

The value is v= 2.73 ft/s to the right  and that of w is 1.85 rad/s counterclockwise.


## Example 18.18-42, Page No 449¶

In [24]:
import math

#Initilization of variables
m1=9 #kg
m2=5.5 #kg
u1=-3 #m/s
u2=1.77 #m/s
e=0.8 #coefficient of restitution

#Calculations
#Solving by matrix method after we get the two equations
A=np.array([[-1,1],[m1,m2]])
B=np.array([[(e*u1-e*u2)],[m1*u1+m2*u2]])
C=np.linalg.solve(A,B) #m/s

#Result
print'The 9kg ball will rebound up the speed of',round(C[0],2),"m/s",'and the 5.5 kg ball will move to the right and down with components of',round(u2,2),"m/s",'and',round(-C[1],2),"m/s respectively."

The 9kg ball will rebound up the speed of 0.26 m/s and the 5.5 kg ball will move to the right and down with components of 1.77 m/s and 3.56 m/s respectively.


## Example 18.18-46, Page No 451¶

In [25]:
import math

#Initilization of variables
v=4 #m/s
m=9 #kg
s=1.5 #m

#Calculations
Io=(2*3**-1)*(m*s**2) #kg.m

#Result
print'The angular velocity of the box is',round(w),"rad/s clockwise."

The angular velocity of the box is 2.0 rad/s clockwise.


## Example 18.18-48, Page No 452¶

In [72]:
import math

#Initilization of variables
m=2000 #kg
mf=8500 #kg
vr=2000 #m/s
a=9.8 #m/s**2

#Calculations
#Applying Newtons Second Law
dm_dt=-(-((m+mf)*a)-(m+mf)*a)/(-vr) #kg/s

#Result
print'dm/dt=',round(dm_dt),"kg/s"
#The negative sign indicates the loss in the mass of the system

dm/dt= -103.0 kg/s


## Example 18.18-50, Page No 453¶

In [74]:
import math

#Initilization of variables
W=4000 #lb
k=3 #ft
d=3.5 #ft
g=32.2 #ft/s**2

#Calculations
I=(W/g)*k**2 #slug-ft**2
M=I*ws*wp #lb-ft
#Now equating M to Rf-Rr gives one equations and vertical sum yields other
#solving them by matrix method
A=np.array([[1,-1],[1,1]])
B=np.array([[M*(2/d)],[W]])
C=np.linalg.solve(A,B) #lb

#Result
print'The weight of Rf and Rr are',round(C[0]),"lb",'and',round(C[1]),"lb respectively."

# The answers in the textbook are incorrect

The weight of Rf and Rr are 3051.0 lb and 949.0 lb respectively.


## Example 18.18-51, Page No 454¶

In [78]:
import math

#As the integration is indefinite we will directly consider the equation with R
#Initillization of variables
GM=1.41*10**16 #ft**3/s**2
r=2640000 #ft
theta=60 #degrees
R=21120000 #ft

#Calculations
v1=sqrt((GM*((R**-1)-((R+r)**-1)))/2.031) #ft/s

#Result
print'The speed required will be',round(v1),"ft/s"

# Answer may wary due to decimal point discrepancy.

The speed required will be 6043.0 ft/s


## Example 18.18-52, Page No 454¶

In [26]:
import math

#Initilization of variables
k=4 #lb/ft
so=1 #ft
W=2**-1 #lb
g=32.2 #ft/s**2
vo=5 #ft/s

#Calculations
m=W/g #kg
#Angular momentum is conserved
v=sqrt((0.5*k*so**2*2*2*g)+vo**2) #ft/s
#Using vd=15
d=15/v #ft

#Result
print'The ball passes',round(d,2),"ft close to the fixed pin"

The ball passes 0.89 ft close to the fixed pin