initial commit / add all books

11 files changed, 218 insertions, 0 deletions

diff --git a/3204/CH18/EX18.1/Ex18_1.sce b/3204/CH18/EX18.1/Ex18_1.sce
new file mode 100644
index 000000000..3f8d861f8
--- /dev/null
+++ b/3204/CH18/EX18.1/Ex18_1.sce
@@ -0,0 +1,15 @@
+// Initilization of variables

+m_a=1 // kg // mass of the ball A

+v_a=2 // m/s // velocity of ball A

+m_b=2 // kg // mass of ball B

+v_b=0 // m/s // ball B at rest

+e=1/2 // coefficient of restitution

+// Calculations

+// Solving eqn's 1 & 2 using matrix for v'_a & v'_b,

+A=[1 2;-1 1]

+B=[2;1]

+C=inv(A)*B

+// Results

+clc

+printf('The velocity of ball A after impact is %f m/s \n',C(1))

+printf('The velocity of ball B after impact is %f m/s \n',C(2))

diff --git a/3204/CH18/EX18.10/Ex18_10.sce b/3204/CH18/EX18.10/Ex18_10.sce
new file mode 100644
index 000000000..3145a54b5
--- /dev/null
+++ b/3204/CH18/EX18.10/Ex18_10.sce
@@ -0,0 +1,17 @@
+// Initilization of variables

+v_a=600 // m/s // velocity of the bullet before impact

+v_b=0 // m/s // velocity of the block before impact

+w_b=0.25 // N // weight of the bullet

+w_wb=50 // N // weight of wodden block

+mu=0.5 // coefficient of friction between the floor and the block

+g=9.81 // m/s^2 // acc due to gravity

+// Calculations

+m_a=w_b/g // kg // mass of the bullet

+m_b=w_wb/g // kg // mass of the block

+// Let the common velocity be v_c which is given by eq'n (Principle of conservation of momentum)

+v_c=(w_b*v_a)/(w_wb+w_b) // m/s

+// Let the distance through which the block is displaced be s, Then s is given by eq'n

+s=v_c^2/(2*g*mu) // m

+// Results

+clc

+printf('The distance through which the block is displaced from its initial position is %f m \n',s)

diff --git a/3204/CH18/EX18.11/Ex18_11.sce b/3204/CH18/EX18.11/Ex18_11.sce
new file mode 100644
index 000000000..b67388180
--- /dev/null
+++ b/3204/CH18/EX18.11/Ex18_11.sce
@@ -0,0 +1,12 @@
+// Initilization of variables

+M=750 // kg // mass of hammer

+m=200 // kg // mass of the pile

+h=1.2 // m // height of fall of the hammer

+delta=0.1 // m // distance upto which the pile is driven into the ground

+g=9.81 // m/s^2 // acc due to gravity

+// Caculations

+// The resistance to penetration to the pile is given by eq'n,

+R=(((M+m)*g)+((M^2*g*h)/((M+m)*delta)))*(10^-3) // kN 

+// Results

+clc

+printf('The resistance to penetration to the pile is %f kN \n',R)

diff --git a/3204/CH18/EX18.2/Ex18_2.sce b/3204/CH18/EX18.2/Ex18_2.sce
new file mode 100644
index 000000000..0100f3a68
--- /dev/null
+++ b/3204/CH18/EX18.2/Ex18_2.sce
@@ -0,0 +1,26 @@
+// Initilization of variables

+m_a=2 // kg // mass of ball A

+m_b=6 // kg // mass of ball B

+m_c=12 // kg // mass of ball C

+v_a=12 // m/s // velocity of ball A

+v_b=4 // m/s // velocity of ball B

+v_c=2 // m/s // velocity of ball C

+e=1 // coefficient of restitution for perfectly elastic body

+// Calculations

+// (A)

+// Solving eq'n 1 & 2 using matrix for v'_a & v'_b,

+A=[2 6;-1 1]

+B=[48;8]

+C=inv(A)*B

+// Calculations

+// (B)

+// Solving eq'ns 3 & 4 simultaneously using matrix for v'_b & v'_c

+P=[1 2;-1 1]

+Q=[12;6]

+R=inv(P)*Q

+// Results (A&B)

+clc

+printf('The velocity of ball A after impact on ball B is %f m/s \n',C(1)) // here the ball of mass 2 kg is bought to rest

+printf('The velocity of ball B after getting impacted by ball A is %f m/s \n',C(2))

+printf('The final velocity of ball B is %f m/s \n',R(1)) // here the ball of mass 6 kg is bought to rest

+printf('The velocity of ball C after getting impacted by ball B is %f m/s \n',R(2))

diff --git a/3204/CH18/EX18.3/Ex18_3.sce b/3204/CH18/EX18.3/Ex18_3.sce
new file mode 100644
index 000000000..34b3f2212
--- /dev/null
+++ b/3204/CH18/EX18.3/Ex18_3.sce
@@ -0,0 +1,13 @@
+// Initilization of variables

+h_1=9 // m // height of first bounce

+h_2=6 // m // height of second bounce

+// Calculations

+// From eq'n (5) we have, Coefficient of restitution between the glass and the floor is,

+e=sqrt(h_2/h_1)

+// From eq'n 3 we get height of drop as,

+h=h_1/e^2 // m

+// Results

+clc

+printf('The ball was dropped from a height of %f m \n',h)

+printf('The coefficient of restitution between the glass and the floor is %f \n',e)

+// Here we use h`=h_1 & h``=h_2 because h` & h`` could not be defined in Scilab.

diff --git a/3204/CH18/EX18.4/Ex18_4.sce b/3204/CH18/EX18.4/Ex18_4.sce
new file mode 100644
index 000000000..18ae94ad8
--- /dev/null
+++ b/3204/CH18/EX18.4/Ex18_4.sce
@@ -0,0 +1,33 @@
+// Initilization of variables

+e=0.90 // coefficient o restitution

+v_a=10 // m/s // velocity of ball A

+v_b=15 // m/s // velocity of ball B

+alpha_1=30 // degree // angle made by v_a with horizontal

+alpha_2=60 // degree // angle made by v_b with horizontal

+// Calculations

+// The components of initial velocity of ball A:

+v_a_x=v_a*cosd(alpha_1) // m/s

+v_a_y=v_a*sind(alpha_1) // m/s

+// The components of initial velocity of ball B:

+v_b_x=-v_b*cosd(alpha_2) // m/s

+v_b_y=v_b*sind(alpha_2) // m/s

+// From eq'n 1 & 2 we get,

+v_ay=v_a_y // m/s // Here, v_ay=(v'_a)_y

+v_by=v_b_y // m/s // Here, v_by=(v'_b)_y

+// On adding eq'n 3 & 4 we get,

+v_bx=((v_a_x+v_b_x)+(-e*(v_b_x-v_a_x)))/2 // m/s // Here. v_bx=(v'_b)_x

+// On substuting the value of v'_b_x in eq'n 3 we get,

+v_ax=(v_a_x+v_b_x)-(v_bx) // m/s // here, v_ax=(v'_a)_x

+// Now the eq'n for resultant velocities of balls A & B after impact are,

+v_A=sqrt(v_ax^2+v_ay^2) // m/s

+v_B=sqrt(v_bx^2+v_by^2) // m/s

+// The direction of the ball after Impact is,

+theta_1=atand(-(v_ay/v_ax)) // degree

+theta_2=atand(v_by/v_bx) // degree

+// Results

+clc

+printf('The velocity of ball A after impact is %f m/s \n',v_A)

+printf('The velocity of ball B after impact is %f m/s \n',v_B)

+printf('The direction of ball A after impact is %f degree \n',theta_1)

+printf('The direction of ball B after impact is %f degree \n',theta_2)

+// Her we use, (1) v'_a & v'_b as v_A & v_B. 

diff --git a/3204/CH18/EX18.5/Ex18_5.sce b/3204/CH18/EX18.5/Ex18_5.sce
new file mode 100644
index 000000000..a71f5d650
--- /dev/null
+++ b/3204/CH18/EX18.5/Ex18_5.sce
@@ -0,0 +1,19 @@
+// Initiization of variables

+theta=30 // degrees // ange made by the ball against the wall

+e=0.50

+// Calculations

+// The notations have been changed

+// Resolving the velocity v as,

+v_x=cosd(theta)

+v_y=sind(theta)

+V_y=v_y

+// from coefficient of restitution reation

+V_x=-e*v_x

+// Resultant velocity

+V=sqrt(V_x^2+V_y^2)

+theta=atand(V_y/(-V_x)) // taking +ve value for V_x

+// NOTE: Here all the terms are multiplied with velocity i.e (v).

+// Results

+clc

+printf('The velocity of the ball is %f v \n',V)

+printf('The direction of the ball is %f degrees \n',theta)

diff --git a/3204/CH18/EX18.6/Ex18_6.sce b/3204/CH18/EX18.6/Ex18_6.sce
new file mode 100644
index 000000000..ada754797
--- /dev/null
+++ b/3204/CH18/EX18.6/Ex18_6.sce
@@ -0,0 +1,12 @@
+// Initilization of variables

+e=0.8 // coefficient of restitution

+g=9.81 // m/s^2 // acc due to gravity

+// Calcuations

+// Squaring eqn's 1 &2 and Solving eqn's 1 & 2 using matrix for the value of h

+A=[-1 (2*g);-1 -(1.28*g)]

+B=[0.945^2;(-0.4*9.81)]

+C=inv(A)*B // m

+// Results

+clc

+printf('The height from which the ball A should be released is %f m \n',C(2))

+// The answer given in the book i.e 0.104 is wrong.

diff --git a/3204/CH18/EX18.7/Ex18_7.sce b/3204/CH18/EX18.7/Ex18_7.sce
new file mode 100644
index 000000000..974032950
--- /dev/null
+++ b/3204/CH18/EX18.7/Ex18_7.sce
@@ -0,0 +1,9 @@
+// Initilization of variables

+theta_a=60 // degree // angle made by sphere A with the verticle

+e=1 // coefficient of restitution for elastic impact

+// Calculations

+// theta_b is given by the eq'n cosd*theta_b=0.875, hence theta_b is,

+theta_b=acosd(0.875) // degree

+// Results

+clc

+printf('The angle through which the sphere B will swing after the impact is %f degree \n',theta_b)

diff --git a/3204/CH18/EX18.8/Ex18_8.sce b/3204/CH18/EX18.8/Ex18_8.sce
new file mode 100644
index 000000000..9edfd8a39
--- /dev/null
+++ b/3204/CH18/EX18.8/Ex18_8.sce
@@ -0,0 +1,39 @@
+// Initilization of variables

+m_a=0.01 // kg // mass of bullet A

+v_a=100 // m/s // velocity of bullet A

+m_b=1 // kg // mass of the bob

+v_b=0 // m/s // velocity of the bob

+l=1 // m // length of the pendulum

+v_r=-20 // m/s // velocity at which the bullet rebounds the surface of the bob // here the notation for v'_a is shown by v_r

+v_e=20 // m/s // velocity at which the bullet escapes through the surface of the bob // here the notation for v_a is shown by v_e

+g=9.81 // m/s^2 // acc due to gravity

+// Calculations

+// Momentum of the bullet & the bob before impact is,

+M=(m_a*v_a)+(m_b*v_b) // kg.m/s......(eq'n 1)

+// The common velocity v_c ( we use v_c insted of v' for notation of common velocity) is given by equating eq'n 1 & eq'n 2 as,

+// (a) When the bullet gets embedded into the bob

+v_c=M/(m_a+m_b) // m/s

+// The height h to which the bob rises is given by eq'n 3 as,

+h_1=(1/2)*(v_c^2/g) // m

+// The angle (theta_1) by which the bob swings corresponding to the value of height h_1 is,

+theta_1=acosd((l-h_1)/l) // degree

+// (b) When the bullet rebounds from the surface of the bob

+// The velocity of the bob after the rebound of the bullet from its surface is given by equating eq'n 1 & eq'n 4 as,

+v_bob_rebound=M-(m_a*v_r) // m/s // here v_bob_rebound=v'_b

+// The equation for the height which the bob attains after impact is,

+h_2=(v_bob_rebound^2)/(2*g) // m

+// The corresponding angle of swing 

+theta_2=acosd((l-h_2)/l) // degree

+// (c) When the bullet pierces and escapes through the bob

+// From eq'n 1 & 5 the velocity attained by the bob after impact is given as,

+v_b_escape=M-(m_a*v_e) // m/s // here we use, v_b_escape insted of v'_b

+// The equation for the height which the bob attains after impact is,

+h_3=(v_b_escape^2)/(2*g) // m

+// The corresponding angle of swing 

+theta_3=acosd((l-h_3)/(l)) // degree

+// Results

+clc

+printf('(a) The maximum angle through which the pendulum swings when the bullet gets embeded into the bob is %f degree \n',theta_1)

+printf('(b) The maximum angle through which the pendulum swings when the bullet rebounds from the surface of the bob is %f degree \n',theta_2)

+printf('(c) The maximum angle through which the pendulum swings when the bullet escapes from other end of the bob the bob is %f degree \n',theta_3)

+// IN THIS SUM WE HAVE USED DIFFERENT NOTATIONS CONSIDERING DIFFERENT CASES BECAUSE IN THE TEXT BOOK WE HAD 3 VARIABLES WITH SAME NOTATION BUT WITH A DIFFERENT VALUE WHICH COULD NOT BE EXECUTED INTO SCILAB.

diff --git a/3204/CH18/EX18.9/Ex18_9.sce b/3204/CH18/EX18.9/Ex18_9.sce
new file mode 100644
index 000000000..bc0f132fc
--- /dev/null
+++ b/3204/CH18/EX18.9/Ex18_9.sce
@@ -0,0 +1,23 @@
+// Initilization of variables

+W_a=50 // N // falling weight

+W_b=50 // N // weight on which W_a falls

+g=9.81 // m/s^2 // acc due to gravity

+m_a=W_a/g // kg // mass of W_a

+m_b=W_b/g // kg // mass of W_b

+k=2*10^3 // N/m // stiffness of spring

+h=0.075 // m // height through which W_a falls

+// The velocity of weight W_a just before the impact and after falling from a height of h is given from the eq'n, ( Principle of conservation of energy)

+v_a=sqrt(2*g*h) // m/s

+// Let the mutual velocity after the impact be v_m (i.e v_m=v'), (by principle of conservation of momentum)

+v_m=(m_a*v_a)/(m_a+m_b) // m/s

+// Initial compression of the spring due to weight W_b is given by,

+delta_st=(W_b/k)*(10^2) // cm

+// Let the total compression of the spring be delta_t, Then delta_t is found by finding the roots from the eq'n........ delta_t^2-0.1*delta_t-0.000003=0. In this eq'n let,

+a=1

+b=-0.1

+c=-0.000003

+delta_t=((-b+(sqrt(b^2-(4*a*c))))/2*a)*(10^2) // cm // we consider the -ve value

+delta=delta_t-delta_st // cm

+// Results

+clc

+printf('The compression of the spring over and above caused by the static action of weight W_a is %f cm \n',delta)