Conservation of Energy, Centre of Mass and Linear Momentum

[Pioneer]

Well here are 5 questions from my text which I've tried doing but can't seem to twist my ahead around to figure them out. I don't know if it's due to the fact that I'm not applying the relevant concepts. Thanks for any help.

(1). A uniform cord of length 25cm and mass 15g is initally stuck to the ceiling. Later, it hangs vertically from the ceiling with only one end still stuck. What is the change in gravitational potrntial energy of the cord with the change in oreintation? (All I know is that this question involves the use of calculus and that's something which I would like someone to help me sort out.)

(2). Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring-loaded gun that is mounted on a table. The target box is a horizonatal distance D = 2.20m from the edge of the table. Bobby compresses the spring 1.10cm but the centre of the marble falls 27.0cm short of the centre of the box. How far should Rhonda compress the spring to score a direct hit, assuming that neither the spring nor the ball encounters friction in the gun?

(3). A cable of a 1800kg elevator cab snaps when the cab is at rest at the first floor, where the cab is 3.7m above a spring of spring constant k = 0.15MN/m. A safety device clamps the cab against guide rails so tha a constant frictional force of 4.4 kN opposes the cab motion.

(a) Find the speed of the cab just before it hits the spring
(b) Find the maximum distance x that the spring is compressed (the frictional force still acts during the compression)
(c) Find the distance that the cab will bounce back u the shaft
(d) Using the conservation of energy, find the approxiamate total distance that the cab wil move before coming to rest. (assuming that the frictional force on the cab is negligible when the cab is stationary)

(4). A metal soda can of uniform composition has a mass of 0.140kg and is 12.0cm tall. This can is filled with 1.31kg of soda then small holes are drilled into the bottom (with negligible loss of metal) to drain the soda. What is the height h of the centre of mass of the can and contents

(a) intially and (b) after the can loses all of the soda
(c) What happens to h as the soda drains?
(d) If X amount remains in the soda at a given instant, what is X when the centre of mass is at its lowest point?

(5). A ball of mass 300g with a speed of 6.0m/s strikes a wall at an angle of 30 degrees and then rebounds with the same speed and angle. It is in contact with the wall for 10 ms. In unit vectors, what are

(a) the impulse on the ball from the wall and
(b) the average force on the wall from the ball?

 

[HallsofIvy]

I don't see any attempt to do the problems on your part! We can't help you if we don't know what you CAN do and what you do or don't understand about the problems.

 

[Pioneer]

Here's what I did for no.5 even though I can't seem to complete it. I truly don't know how to tackle the other problems though.

So here I go

J = pf - pi = m(vf-vi)
where pf = the momentum just after the ball strike the wall for 10ms
How do i find what vf is?

Jx =m(vfx - vix)
=.3[vfx*cos(-30) - 6*cos30)]

Jy = .3[vfy*sin(-30) - 6*sin30)]

therefore

J = Jxî - Jyĵ kg.m/s

 

[lightgrav]

Question 1 involves the use of gravitational PE "formula" mgh, not calculus.

1) draw a sketch of the situation.
2) most Energy and momentum scenarios involve TWO situations, at least,
so it's usually best to draw 2 sketches, so you can see what's different.
Describe in words what's happening. Describe what the Energy does, or what happens to the motion.

You DO have to learn to read with comprehension - this is a start.

final velocity?

rebounds with the same speed and angle

 

[BerryBoy]

Why $$-30^\circ$$?

I agree that you have to draw this out (at absolute least) once. Then - might become + ...

You may want to look at the DIRECTION of the velocity in the X direction, how would this modify your calculation?

Regards,
Sam

 

[Pioneer]

okay, let me do this again

Jx = .3[6cos30] - .3[cos30] = 0
Jy = .3[6sin30] - .3[6sin(-30)] = (1.8N.s)ĵ

Therefore F = 180/0.01 = (180N)ĵ

 

[dx]

Notice that the y component of the velocity doesn't not change. That means that the force is normal to the wall . What is the change in the x component of velocity?

$$\frac {F}{m} \Delta{t} = \Delta{v}$$

 

[Pioneer]

Alright here's a next question which I've attempted nd would like a bit of help with.

I drew a diagram already so here's the question
A cart is conneted by a cord to a hanging block. The cart has a mass m1 = 0.6kg and it's centre of mass is intially at xy coordiantes (-.500m, 0m); the block has a mass m2 = 0..400kg and its centre is intially at xy coordinates (0m, -0.100m). The mass of the cord and the pulley are negligible. The cart is released from rest and both the cart and the block move until the cart hits the pulley. The friction between the cart and the air track and the pulley and its axle is negligible.
(a) In unit vector notation, what is the acceleration of the mass-cart system
(b)What is the velocity of the com as a function off time t?

This is what I've done so far
Forces on m1 = m1g
Forces on m2 = m2g and T

therefore m1g = ma1
T - m2g = ma2

What do I do from there?

For (b)
acom = dvcom/dt
therefore vcom = acomtf - acomti