Recent content by pmrazavi

You should use series method to solve this diff eq. It's hard to explain the method in this tiny box(!). I suggest to check out an elemantary diff eq book and read the chapter about the series method.

It's just about free body diagrams...draw the free body diagram for both of the masses..then apply Newton's second law: for the hanging mass you will get m1g-T=m1a and for the block on the table you will get 2T=m2a..add up these 2 equations you will get T=(m1+m2)a-m1g..this is the tension in...

YES, of course the compression will be diffrent because of the gravity's force... In this situation you don't only have kinetic energy but you have the gravity potential energy as well. The equation will be this then .5kx^2=.5mV2+mgh (in the vertical bouncing)... but if it ball is traveling a...

I have no other idea..b/c I'm sure my anwer is correct :)...maybe the book's answer is wrong or the book's given numbers are not the same as you've posted... or maybe it wants the answer in jouls not ft*lb... Notice that the work of gravity is actually mgh, which "h" is the height the object...

in the equation: .5kx^2=mgh, notice that both of these energies are not for the initial or the final point, the left side is the energy of the start point that we only got spring, and the right side is mgh, which is in the final place and there is no spring... i hope I've understood your...

when she is about to lose contact, the Normal force is zero. At the top of the cliff the only type of energy is Pot. grav. (mgh), but when she is about to lose contact she has both potential and kinetic energy which their sum is equal to the first mgh. You should find the angle by computing...

Gravity's work is -mgsin30d in this case. In incline problems gravity has 2 components: 1) y-component, mgcos30, which doesn't do work (perpendicular), and the other one is mgsin30. W=F.d (dot prodoct) or W=Fdcos (theta). ==> work of this force is mgsin30d(cos180), which is -mgdsin30..

You should assume all of the collisions to be inelastic, unless it's said that it's elastic.

ONE OF THE PHYSICS LAW, CONSERVATION OF MOMENTUM: Momentum in a system is always conserved: m1v1i+m2v2I=m1v1f+m2v2f No, not necessarily. B/c we have conservation of MOMENTUM, it means that sum of the products of mass and velocity in a system, containing any number of objects, is...

by the way..you can solve it without using energy stuff: use vf^2-Vi^2=2a(Xf-Xi) You got the delta X, find "a" by F=ma, finalV=0 so you can easily find Vi..

I'm really curious to know what you got from the exam.. :)

the final answer for Vi=273.28 m/s

you should find the "delta X", by this equation: mgh=.5kx^2, I got 30.45 m for delta X, then you should put this value in the equation: mgh=.5kx^2+mgh`+.5mv^2 and find the velocity..the initial X, 10.1, is extra in the given problem.

The surface MUST have friction, else it would continue going on with the final velocity...I guess you should look up the friction coeffiecent of wood, then solve the problem... m1v1=(m1+m2)Vf (.001)V1=(10.001)Vf You are looking for V1..to do that first you should find Vf: you should...