8 files changed, 107 insertions, 0 deletions
diff --git a/1205/CH14/EX14.1/S_14_1.sce b/1205/CH14/EX14.1/S_14_1.sce
new file mode 100644
index 000000000..058806112
--- /dev/null
+++ b/1205/CH14/EX14.1/S_14_1.sce
@@ -0,0 +1,20 @@
+clc;
+m=200;//kg, mass of space vehicle
+vo=150;//m/s i, relative velocty of vehicle with frame at t=0
+mA=100;//kg, mass of part A
+mB=60;//kg, mass of part B
+mC=40;//kg, mass of part C
+
+t=2.5;//s, given time
+A=[555,-180,240];//m, Position of A at t=2.5
+B=[255,0,-120];//m, Position of B at t=2.5
+
+r=[vo*t,0,0];//m,  Position of mass center G
+
+//recalling 14.12
+//m*r=mA*rA+mB*rB+mC*rC
+rA=A;//m, 
+rB=B;//m
+
+rC=(m*r-mA*rA-mB*rB)/mC;//m, Position of part c at t=2.5 s
+printf("Position of part c at t=2.5 s is rC= (%.0f m)i +(%.0f m)j +(%.0f m)k\n",rC(1),rC(2),rC(3));
diff --git a/1205/CH14/EX14.2/S_14_2.sce b/1205/CH14/EX14.2/S_14_2.sce
new file mode 100644
index 000000000..b39f21f39
--- /dev/null
+++ b/1205/CH14/EX14.2/S_14_2.sce
@@ -0,0 +1,18 @@
+clc;
+m=10;//kg, mass of projectile before exploding
+v=30;//m/s, velocity of projectile before exploding
+mA=2.5;//kg, mass of fragment A of projectile
+mB=7.5;//kg, mass of fragment B of projectile
+thetaA=45;//degree, direction of fragment A
+thetaA=thetaA*%pi/180;//rad, conversion into radian
+thetaB=30;//degree, direction of fragment B
+thetaB=thetaB*%pi/180;//rad, conversion into radian
+//Law of conservation of momentum gives
+//mA*vA+mB*vB=m*v
+//Taking x and y components of vA and vB we get
+A=[mA*cos(thetaA),mB*cos(thetaB);mA*sin(thetaA),-mB*sin(thetaB)];// matrix of coefficients
+B=[m*v;0];//matrix B, v is along X axis
+X=linsolve(A,-B);//solution matrix
+vA=X(1);//m/s, velocity of fragment A
+vB=X(2);//m/s, velocity of fragment B
+printf("Velocity of fragment A after explosion is %.2f m/s \n Velocity of fragment B after explosion is %.1f m/s with given directions\n",vA,vB);
diff --git a/1205/CH14/EX14.3/S_14_3.sce b/1205/CH14/EX14.3/S_14_3.sce
new file mode 100644
index 000000000..f415e7249
--- /dev/null
+++ b/1205/CH14/EX14.3/S_14_3.sce
@@ -0,0 +1,26 @@
+clc;
+m=200;//kg, mass of space vehicle
+vo=150;//m/s i, relative velocty of vehicle with frame at t=0
+mA=100;//kg, mass of part A
+mB=60;//kg, mass of part B
+mC=40;//kg, mass of part C
+vA=[270,-120,160];//m/s, velocity of A
+t=2.5;//s, given time
+A=[555,-180,240];//m, Position of A at t=2.5
+B=[255,0,-120];//m, Position of B at t=2.5
+
+r=[vo*t,0,0];//m,  Position of mass center G
+
+//recalling 14.12
+//m*r=mA*rA+mB*rB+mC*rC
+rA=A;//m, 
+rB=B;//m
+
+rC=(m*r-mA*rA-mB*rB)/mC;//m, Position of part c at t=2.5 s
+//m*vo=mA*vA+mB*vB+mC*vC and   1
+//Ho1=(Ho)2 ....2  gives equation 
+// Equate i coefficient of 1 and j and k coefficients of 2 to zero as B lies in xz plane
+vCy=300;//m/s, y component of velocity of part C
+vCz=-420*vCy/450;//m/s, z component of velocity of part C
+vCx=(105*vCy-45000)/450;//m/s, x component of velocity of part C
+printf("Velocity of part c at t=2.5 s is vC= (%.0f m)i +(%.0f m)j +(%.0f m)k\n",vCx,vCy,vCz);
diff --git a/1205/CH14/EX14.4/S_14_4.sce b/1205/CH14/EX14.4/S_14_4.sce
new file mode 100644
index 000000000..74eb56bac
--- /dev/null
+++ b/1205/CH14/EX14.4/S_14_4.sce
@@ -0,0 +1,3 @@
+
+clc;
+printf("Given problem is theoritical and no mathematical solving required for this problem");
diff --git a/1205/CH14/EX14.5/S_14_5.sce b/1205/CH14/EX14.5/S_14_5.sce
new file mode 100644
index 000000000..eae86f93a
--- /dev/null
+++ b/1205/CH14/EX14.5/S_14_5.sce
@@ -0,0 +1,16 @@
+clc;
+//By theoritical work, applying law of conservation of momentum and energy we get
+//vA=(vB)y=3vc-6, (vB)x=3-vc
+//20*vc^2-78*vc+72=0
+
+y=poly([72,-78,20],'x','coeff');//Obtained polynomial
+vc=roots(y);//m/s, 
+vc=vc(1);//AS vc(2) gives negative value of vA
+vA=3*vc-6;// m/s Velocity with which ball A hits the side of table
+vBy=3*vc-6;//m/s, y coefficient Velocity with which ball B hits the side of table
+vBx=3-vc;//m/s, x coefficient Velocity with which ball B hits the side of table
+vB=[vBx,vBy];//m/s Velocity with which ball B hits the side of table
+theta=atan(-vBy/vBx);//rad, angle of velocity B
+theta=theta*180/%pi;//degree
+
+printf("Velocities with which balls  hits the sides of table are\n vA= %.1f m/s \n vB= %.3f m/s with theta= %.1f degree \n vC=%.1f m/s\n",vA,norm(vB),theta,vc);
diff --git a/1205/CH14/EX14.6/S_14_6.sce b/1205/CH14/EX14.6/S_14_6.sce
new file mode 100644
index 000000000..9bc37ff34
--- /dev/null
+++ b/1205/CH14/EX14.6/S_14_6.sce
@@ -0,0 +1,18 @@
+clc;
+W=3000;//N, force applied at G
+delmt=120;//kg/s, rate of falling grains
+vA=10;//m/s, velocity with which grains hits chute at A
+vB=7.5;//m/s, velocity with which grains hits chute at B
+theta=10;//degree,angle with which grains falls
+theta=theta*%pi/180;//rad
+//System formed by the momentum and impulses is equipollent to the momentum del_m*vB
+
+//putting above values in equations we get
+
+B=(3.5*W+1.5*delmt*vA+3*delmt*vB*cos(theta)-6*delmt*vB*sin(theta))/6;//N, reaction at roller support B
+Cx=delmt*vB*cos(theta);//N, x component of reaction at hinge C
+Cy=W-B+delmt*(vA-vB*sin(theta));//N, x component of reaction at hinge C
+
+printf("Reaction at roller support B is B= %.0f N \n x component of reaction at hinge C is Cx= %.0f N\n  y component of reaction at hinge C Cy=%.0f N\n",B,Cx,Cy);
+
+//Actual solving also shows similar answer as programme shows, book shows rounded value
diff --git a/1205/CH14/EX14.7/S_14_7.sce b/1205/CH14/EX14.7/S_14_7.sce
new file mode 100644
index 000000000..74eb56bac
--- /dev/null
+++ b/1205/CH14/EX14.7/S_14_7.sce
@@ -0,0 +1,3 @@
+
+clc;
+printf("Given problem is theoritical and no mathematical solving required for this problem");
diff --git a/1205/CH14/EX14.8/S_14_8.sce b/1205/CH14/EX14.8/S_14_8.sce
new file mode 100644
index 000000000..74eb56bac
--- /dev/null
+++ b/1205/CH14/EX14.8/S_14_8.sce
@@ -0,0 +1,3 @@
+
+clc;
+printf("Given problem is theoritical and no mathematical solving required for this problem");