initial commit / add all books

1 files changed, 42 insertions, 0 deletions

diff --git a/1325/CH2/EX2.2/ex2_2.sce b/1325/CH2/EX2.2/ex2_2.sce
new file mode 100644
index 000000000..d4c24527f
--- /dev/null
+++ b/1325/CH2/EX2.2/ex2_2.sce
@@ -0,0 +1,42 @@
+//to find speed of truck immidiately after collision and the maximum deflection of spring during impact. Moreover if k=0.5 then determine hoe the final speedsw will be affected and amount of dissipated energy 

+clc

+//given

+m1=15//tons

+u1=12//m/h

+m2=5//tons

+u2=8//m/h

+k=2//ton/in

+e1=0.5//coefficient of restitution

+printf("\n")

+//conservation of linear momentum

+v=(m1*u1+m2*u2)/(m1+m2)

+printf("velocity at the instant of  collision is %.2f mph",v)

+e=(m1*m2*(88/60)^2*(u1-u2)^2)/(2*32.2*(u1+u2))

+printf("\n")

+printf("The difference between the kinetic energy before and during the impact is %.2f ft tons\n",e)

+//energy stored in spring equals energy dissipated

+//s=(1/2)*k*x^2

+//s=e

+//since there are 4 buffer springs ,4x^2=24 inches (2 ft=24 inches)

+x=((e*12)/4)^.5

+printf("Maximum deflection of the spring is %.2f in\n",x)

+// maximum force acting between pair of buffer = stiffness of spring*deflection

+f=k*x

+printf("Maximum force acting between each buffer is %.2f tons\n",f)

+//assuming perfectly elastic collision

+//for loaded truck 

+v1=2*11-12

+//for unloaded truck 

+v2=2*11-8

+printf("Speed of loaded truck after impact %.2f mph\n",v1)

+printf("speed of unloaded truck after impact %.2f mph\n",v2)

+//if coefficient of restitution =o.5

+//for loaded truck 

+ve1=(1+.5)*11-.5*12

+//for unloaded truck 

+ve2=(1+.5)*11-.5*8

+printf("Speed of loaded truck after impact when e=0.5 %.2f mph\n",ve1)

+printf("Speed of unloaded truck after impact when e=0.5 %.2f mph\n",ve2)

+//net loss of kinetic energy=(1-e^2)*energy stored in spring

+l=(1-(e1^2))*2//ft tons

+printf("Net loss of kinetic energy is %.2f ft tons\n",l)