# CHAPTER09 : CENTRE OF MASS LINEAR MOMENTUM COLLISION¶

## Example E4 : Pg 146¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 9.4
# calculation of the maximum compression of the string
# given data
import math
m=1.# mass(in kg)
v=2.# speed of the block(in m/s)
k=50.# spring constant(in N/m)
# calculation
V=(m*v)/(m+m)# principle of conservation of linear momentum
ke1=(m*v*v/2.)# initial kinetic energy
ke2=(m*V*V/2.)+(m*V*V/2.)# final kinetic energy
x=math.sqrt(2.*(ke1-ke2)/k)# kinetic energy lost = elastic energy stored
print'the maximum compression of the string is',x,'m'

the maximum compression of the string is 0.2 m


## Example E5 : Pg 148¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 9.5
# calculation of the speed of combined mass
# given data
ma=50.# mass(in kg) of cart A
mb=20.# mass(in kg) of cart B
va=20.# velocity(in km/hr) of cart A
vb=10.# velocity(in km/hr) of cart B

# calculation
V=((ma*va)-(mb*vb))/(ma+mb)# principle of conservation of linear momentum

print'the speed of combined mass after collision is',round(V,2),'km/hr'

the speed of combined mass after collision is 11.43 km/hr


# WORKED EXAMPLES¶

## Example E1w : Pg 149¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 9.1w
# Locating the centre of maass of the system

# given data
m1=.50# mass(in kg) at point1
m2=1.# mass(in kg) at point2
m3=1.5# mass(in kg) at point3
x1=0# x coodinate (in cm) of point1
x2=4.# x coodinate (in cm) of point2
x3=0# x coodinate (in cm) of point3
y1=0# y coodinate (in cm) of point1
y2=0# y coodinate (in cm) of point2
y3=3.# y coodinate (in cm) of point3

# calculation
X=((m1*x1)+(m2*x2)+(m3*x3))/(m1+m2+m3)
Y=((m1*y1)+(m2*y2)+(m3*y3))/(m1+m2+m3)

print'The centre of mass is',round(X,2),'cm right and',Y,'cm left above the .5 kg paticle'

The centre of mass is 1.33 cm right and 1.5 cm left above the .5 kg paticle


## Example E6w : Pg 151¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 9.6w
# calculation of the acceleration of the centre of mass
# given data
M=2.5# mass(in kg) of the body
F1=6.# force(in N) acting at point 1
F2=5.# force(in N) acting at point 2
F3=6.# force(in N) acting at point 3
F4=4.# force(in N) acting at point 4
theta1=0# angle(in degree)
theta2=37.# angle(in degree)
theta3=53.# angle(in degree)
theta4=60.# angle(in degree)
# calculation
Fx=3.6;#(-F1*cosd(theta1))+(F2*cosd(theta2))+(F3*cosd(theta3))+(F4*cosd(theta4))# X component of resultant force
Fy=1.68;#(F1*sind(theta1))+(F2*sind(theta2))+(-F3*sind(theta3))+(F4*sind(theta4))# X component of resultant force
F=3.98;#sqrt((Fx*Fx)+(Fy*Fy))
theta=25.;#atand(Fy/Fx)
acm=1.59;#F/M# acceleration of centre of mass
print'the acceleration of the centre of mass is',acm,'m/s**2 and is in the direction of the resultant force'

the acceleration of the centre of mass is 1.59 m/s**2 and is in the direction of the resultant force


## Example E8w : Pg 151¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 9.8w
# calculation of the distance from launching point

# given data
u=100.# speed(in m/s) of the projectile
theta=37.# angle(in degree) of the projectile above horizontal
g=10.# gravitational acceleration(in m/s**2) of the earth

# calculation
xcm=961.;#(2.*u*u*sind(theta)*cosd(theta))/g# range of original projectile
# also   xcm=((m1*x1)+(m2*x2))/(m1+m2)
# here m1=M/4   and    m2=3*M/4
x1=xcm/2.# since small part falls from heighest point i.e half of range
x2=(4./3.)*((xcm*((1./4.)+(3./4.)))-(x1/4.))

print'the distance of landing of heavier piece from launching point is',round(x2,2),'m'

the distance of landing of heavier piece from launching point is 1121.17 m


## Example E9w : Pg 152¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 9.9w
# calculation of the distance moved by the bigger block
# given data
L=2.2# length(in m) of the base
n=10.#  mass of bigger block is 'n' number of times the mass of smaller block
# calculation
# centre of mass at rest initially will remain in horizontal position thus
# M*(L-X)=10*M*X
X=L/(n+1.)
print'distance moved by the bigger block at the instant the smaller block reaches the ground is',X,'m'

distance moved by the bigger block at the instant the smaller block reaches the ground is 0.2 m


## Example E10w : Pg 152¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 9.10w
# calculation of the average force exerted by the hero on the machine gun

# given data
m=50.*10.**-3.# mass(in kg) of the bullet
v=1.*10.**3.# velocity(in m/s) of the bullet
n=20.# number of bullets fired
t=4.# time(in s) required in firing the bullets

# calculation
me=m*v# momentumof each bullet
f=me*n/t# force=rate of change of momentum

print'the average force exerted by the hero on the machine gun is',f,'N'

the average force exerted by the hero on the machine gun is 250.0 N


## Example E11w : Pg 152¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 9.11w
# calculation of the fractional change in kinetic energy

# given data
vb=20.# speed(in m/s) of the block
v1=30.# velocity(in m/s) of one of the part

# calculation
M=1.# taking mass M=1 kg  for solving the equation
v=(1./M)*((M*vb*2.)-(M*v1))# principle of conservation of linear momentum
deltake=(M*v1*v1/(2.*2.))+(M*v*v/(2.*2.))-(M*vb*vb/2.)# change in the kinetic energy
fdeltake=deltake/(M*vb*vb/2)# fractional change in the kinetic energy

print'the fractional change in the kinetic energy is',fdeltake

the fractional change in the kinetic energy is 0.25


## Example E13w : Pg 152¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 9.13w
# calculation of the final velocity of the shuttle

# given data
v1=4000.# speed(in km/hr) of shuttle with respect to the earth
v2=100.# speed(in km/hr) of the module with respect to the shuttle

# calculation
M=1.# taking mass M=1 kg  for solving the equation
vdash=v1-v2# speed of module with respect to the earth
V=(1./5.)*((1.*v1*6.)-(vdash*1.))# principle of conservation of linear momentum

print'the final velocity of the shuttle is',V,'km/h'

the final velocity of the shuttle is 4020.0 km/h


## Example E14w : Pg 153¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 9.14w
# calculation of the velocity with which the board recoils

# given data
m1=25.# mass(in kg) of the boy
m2=10.# mass(in kg) of the board
v1=5.# velocity(in m/s) of the boy

# calculation
v=(m1*v1)/m2# principle of conservation of linear momentum
vsep=v1+v# velocity of separation

print'the velocity with which the board recoils is',v,'m/s'
print'the velocity of separation of the boy and the board is',vsep,'m/s'

the velocity with which the board recoils is 12.5 m/s
the velocity of separation of the boy and the board is 17.5 m/s


## Example E17w : Pg 153¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 9.17w
# calculation of the speed of the bullet
# given data
import math
mb=50.*10.**-3.# mass(in kg) of the bullet
mp=450.*10.**-3.# mass(in kg) of the bob
h=1.8# height(in m) attained by the bob
g=10.# gravitational acceleration(in m/s**2) of the earth
# calculation
# using principle of conservation of linear momentum  and equation of motion (v*v) = (u*u) + (2*a*x)
v=((mb+mp)*(math.sqrt(h*2.*g)))/mb

print'the speed of the bullet is',v,'m/s'

the speed of the bullet is 60.0 m/s


## Example E22w : Pg 155¶

In :
# developed in windows XP operating system 32bit
# platform Scilab 5.4.1
#clc;clear;
# example 9.22w
# calculation of the loss of kinetic energy due to the collision

# given data
m=1.2# mass(in kg) of the block1
v=20.*10.**-2.# velocity(in m/s) of the approach
e=3./5.# value of coefficient of restitution
vdash=e*v# velocity (in m/s) of the separation

# calculation
# by principle of conservation of linear momentum ....v1 + v2 = v m/s.....(1)
# as the coefficient of restitution is 3/5............v2 - v1 = vdash m/s.....(2)
# from equation (1),we get.......v2=v-v1
# substituting v2 in equation (2),we get
v1=(v-vdash)/2.
v2=v-v1# from equation (1)
lke=(m/2.)*((v*v)-(v1*v1)-(v2*v2))

print'the loss of kinetic energy during the collision is',lke,'J'

the loss of kinetic energy during the collision is 0.00768 J