A Little Help on Physics Please!

They are all pretty basic questions, rather than us doing it all for you, why not show us what you've tried so far and we could point our ways to improve your understanding? If we do it all for you what will you learn?

As a starter there are only a very few equations you have to consider here and apply, and most will be solved with just one or at most two of these equations:

1. Conservation of kinetic energy (1/2mv^2) before and after a interaction/collision
2. Conservation of momentum (mv) before and after a collision/interaction
3. Work = force * distance

also you might use s = ut + 1/2 at^2 and v^2 - u^2 = 2 as

haha thanks for the reply

these questions are part of a winter vacation extra credit packet. i've nailed a good protion of the problems (20 in total) but some parts of physics i just completely fail at (or have forgotten).

for #7 i think i have the answer.

F(.08) = 1000(14)
so F=1.75 x 10^5??

im not too sure but i think that's right.

7 looks good.

Looking at 10

Presume it is in the presence of a gravitation field, say Earth at the equator at sea level, so g is 9.81 m/sec/sec. Also presume the numbers on the objects attached to the pulleys are weights stated in kilograms and that the cord connecting them is a strong, inelastic (at that load), massless, flexible cable (i.e. use a hardened 1" steel cord and that system ain't going to move, it'll just pivot to your right).

I'd expect you answers would be anti-clockwise, g and (I think) 3g.

The last part is tricky as you'd expect the tension to be a total of 8g if the system was static, but once it becomes frictionlessly dynamic the force diagram changes.

Imagine that was a 100kg and a 0kg weight - then there would absolutely no tension in the cable as it free fell.

Now think of a 100kg vs a 1kg weight. The 1 kg weight will move upwards but only at 1 g, not 99g! So the tension on the cable would be twice the rest mass to not only hold it against gravity (1 g) but lift it at 1g, divided by 2 (?) as its a pulley with 2 cables so the tension is halved. So tension = twice weight lifted divided by 2 for a two cable pulley.

So in your experiment I think you are lifting a 3kg mass at 1g (=3 Netwons tension) + lightening a 5kg mass to 2kg in a field of g strength (another 3 Newtons) = 6 Newtons (halved by the two pulley system).

Please check this pulleys weren't my thing - bridges not collapsing where!

Question 14.

Assume the boys release their force in a purely forwards direction, not up and forwards at angle theta to the horizon, so you don't have to reduce forces to the cosine of the theta as cos 0 = 1.

Then you just have two momentums (p = mv) to counter balance assuming the boat does crush, and the net momentum must transfer back to the boat weight 100 kgs.

So v boat (horizontal) = (mass 1 * v1 - mass 2 * v2) / mass boat ... (plus in real life the vertical velocity component given by buoyancy as the two lads exit the boat), hope you get 2.7 m/sec away from the heavier lad.

Question 15

Conserve both momentum and energy (magnitude and direction) before and after your collisions. So sum of

i) m1v1 + m2v2 (as vectors, magnitude and direction) = (m1+m2) v new (magnitude and direction
i) 1/2m1 v1^2 + 1/2 m2v2^2 = 1/2(m1 + m2) v new^2

Use sin and cosine to sum the horizontal and vertical force components

Question 16

Sum forces like vectors as above. Angular momentum (a force) = m v^2 / r

Question 19

Assume the ladder is rigid and neither the ground or building have any flex (so you can ignore turning moments and greatly simplify the question). Then when a system is in equilibrium all its forces must be balanced.

So force down is mg somewhere on the ladder (that somewhere is important if turning moments and tension in the ladder are important or non linear).

But for this level physics you have only 4 forces, gravity * mass down, ground pushing up at strength sin angle * mg, wall pushing out at force cosine angle * mg and tension within the ladder replying these three forces with net equal and opposite forces spread throughout the beam.

I use six forces to actually simplify

1. force of ladder on ground = mg sin x
2. force of ground on ladder = - mg sin x (opposite direction)

3. force of ladder on wall = mg cos x
4. force of ladder on wall = - mg cos x

5. force of person on ladder = mg
6. force of ladder on person = -mg