1. Such a collision is called a perfectly inelastic collision because the objects move off together after the collision.
Using the principle of conservation of momentum, you'll find
Momentum before = Momentum after
(5x1010)(7200) + (6x1024)(0) = (6x1024)v
where 'v' is the common velocity (imparted). You can neglect the mass of the meteor on the RHS of the above equation because it is very small compared to the mass of the earth.
Solving for v you'll find
v = 6.4 x 10-11 ms-1
which is very small indeed, as you might have guessed!
2. (i)You must first know what is meant by the law of conservation of momentum. You'll find a statement along with the solution to the problem. It would help you to know that Newton's third law that is popularly known as every action has an equal and opposite reaction actually follows from this momentum conservation and vice versa.
When a bullet is fired it moves on with a large momentum. Initially the momentum of the bullet AND the gun were zero. If the bullet has moved away, then because the total momentum must continue to remain zero, the gun must recoil (move backward) with the same MOMENTUM although with a much smaller velocity thanks to its large mass.
Check out problem (ii) below. This would clarify the ideas further.
The logic is the same with a rocket launch too. The total momentum of the hot gases AND the rocket at the beginning is zero. Therefore when the hot gases are forced at the back carrying a large downward momentum, the rocket must automatically be forced up with THE SAME momentum UPWARDS. The rate at which the hot gases come out helps exert a large upward force on the rocket.
I would also like to suggest some URL's for you to browse and understand this further.
http://www.lerc.nasa.gov/WWW/K-12/airplane/newton3.html
http://www.iit.edu/~smile/ph9408.html
http://www.iit.edu/~smile/ph9608.html
http://www.glenbrook.k12.il.us/gbssci/phys/Class/newtlaws/u2l4a.html
You may find it useful to read through my answer to the question about Newton's third law in the link 'Newton's Laws of Motion' among these Physics Topics.
(ii) This classic problem in 'recoil' involves the principle of conservation of momentum. In the absence of an external force, the total momentum of the system before collision must equal the total momentum after collision.
Initially the total momentum is zero.
Let the mass of the gun be M and the mass of the bullet be m.
Also, let the velocity of the gun be V and the bullet be v.
We can now write
0 = MV + mv
Making v the subject of the equation, you'll have
v = - 1000 ms-1
That is the bullet travels at a speed of 1000 ms-1 opposite to the direction in which the gun moves!
3. This problem is quite common in physics and several students CARELESSLY get it wrong. The reason, they forget velocity is a VECTOR, it has magnitude AND direction.
If you choose the direction of velocity of approach as negative then the direction when it bounces would be POSITIVE. Therefore the change in velocity would be final velocity - initial velocity = [2 - (-3)] = 5 ms-1
Therefore the momentum change of the ball would be
Change = 1(2) - [1(-3)]
= 5 kgms-1 or 5 Ns
4. The concept is simple.
Impulse = change in momentum
Here the velocity of approach is 6 ms-1 and the velocity of rebound is -4 ms-1. The change is velocity is 6 -(-4) = 10 ms-1.
Therefore the impulse = m(v - u)
= 0.2 (10)
= 2 Ns
which is the same if you had used the other mass, because
Impule = m(v - u)
= 1(2 - 0)
= 2 Ns
This is just as expected in an elastic collision. If the collisions were inelastic, then impulse would have been smaller because the final velocity would have been lower on account of loss of energy.
5. This is a problem which involves energy and momentum conservation and you'll find similar problem solved elsewhere in the list of questions answered.
Draw a rough sketch. Using the principle of conservation of energy, you'll have
mgh = 1/2 mv2
Using this you can find the speed of the ball just BEFORE the collision. This gives you
v2 = 2gh
You're also told that this collision is perfectly elastic. This means the total momentum and energy of the system before and after the collision must be the same.
Using the momentum conservation
(0.5)(3.7) + 0 = (0.5)b + (2.5)k
where 'b' is the speed of the ball just after the collision and 'k' is the speed of the block just after the collision.
This equation has two variables. To solve it you need the second equation. Using kinetic energy conservation you'll have
1/2(0.5)(3.7)2 + 0 = 1/2(0.5)b2 + 1/2(2.5)(k)2
Solving the two equation above you must find
b = - 2.45 ms-1 (speed of ball) and
k = + 1.23 ms-1.(speed of block)
That means the ball is going to rebound while the block moves forward.
6. This problem involves the principle of conservation of momentum (PoCoM) and energy. As usual draw a rough sketch to imagine this collision.
Applying the PoCoM first you'll have
m(v) - 300(v) = m(x) + 0
Giving you the equation
x = v(1-300/m)
Since the collision is elastic, you can now apply the energy conservation equation. This would give you
1/2mv2 + 1/2(300)v2 = 1/2m(x)2 + 0
Substitution for 'x' the speed of the second titanium sphere from above momentum equation you'll have
mv2 + 300v2 = mv2[1-300/m]2
having cancelled the constant 1/2 on both sides.
Now v2 is also common and can be cancelled here. Therefore
m + 300 = m[1-300/m]2
Open the square bracket and write
m + 300 = m[1 + 90000/m2 - 600/m]
where you've used (a+b)2 = a2 + b2 + 2ab
Multiplying 'm' inside you'll find
m + 300 = m + 90000/m - 600
The solution of this is obviously
m = 100 g
This is the mass of the other sphere you were looking for.
(b) To find the velocity of the center of mass of the system use the formula
vcm = total linear momentum/total mass
= (100)(2) + 300 (-2)/[100 + 300]
= 200 - 600/400
= - 1 ms-1
This is the constant common velocity of the center of the two titanium spheres. Note the negative sign merely shows it is in the direction of the 300 g sphere,which we have assumed to be moving to the left in our problem.
7. I'd strongly recommend that you study the numerous other problems I've solved on this topic of 'PoCoM' that you'd find among my other answers.
I'll briefly outline the equations here. Like always draw a sketch and work according to that. You have to apply the PoCoM along two prependicular directions.
Considering the x-direction, you'll have
(1.5 x 105)6.2 + 0 = (1.5 x 105)vcosA + (2.78 x 105)5.1sin18
where v is the speed of the first barge immediately after the collision and A is the direction of motion of the first barge (below the horizontal) immediately after the collision.
NB: You must have resolved the speed of the second barge along the horizontal and vertical.
Continuing now along the vertical direction, you'll have
0 + (2.78 x 105)4.3 = (2.78 x 105)5.1cos18 + (1.5 x 105)vsinA
The two equations would reduce to
vsinA = 1.02
and
vcosA = 3.28
Dividing the two equations we have
tanA = 0.311 or A = 17.3o
and
substituting in any one of the equations you'll find
v = 3.43 ms-1
The second part of the problem is straightforward. You find the total kinetic energy using the equation 1/2mv2.
You'll find the energy before = 5.45 x 106 J
and the energy after is = 4.5 x 106. The kinetic energy lost in the collision would be approximately 9.5 x 105 J
8. The shorter the time an object has, greater the force experienced by it. According to Newton's second law, the force is proportional to the rate of change of momentum. The momentum change is the same in case of hard and soft objects but by having a shorter time of impact, the hard object experiences and exerts in its turn a greater force.
It is not true that you experience no force when you undergo free fall. As you speed up the air resistance too increases on you. Otherwise you can expect to reach very very high velocities on reaching the ground, which is not the case anyway.
A parachute can only help apply a large drag force for a short time and slow you down significantly. Your weight will in any case pull you down and there is no way you're going to float. That was man's greatest dream before he could actually fly and it is still a facination for most people to see planes fly at ease, almost as free as a bird!
If the force is greater than what an object can take it is either deformed or breaks.
The reason some insects are able to float on water is because of the force acting on the surface of water called SURFACE TENSION, something like a skin being stretched on water. Why insects, if you even try and fload a needle with a filter paper it floats although steel is almost 8 times as dense as water! If you want a detalied explanation for surface tension do let me know.
9. You have to understand forces between molecules first to explain surface tension. Instead of drawing a graph, I'm going to ask you to think about these principles. All atoms and objects would like to be in a state of least potential energy.
The separation of the molecules at this energy is called the equilibrium separation. If the molecules are pushed closer together, they'll experience a force of repulsion. If they're pulled farther apart from the equilibrium separation, then they'll experience an attractive force.
This is exactly what happens at the surface of a liquid. Since you don't have molecules above the surface, the molecules at the top experience a net downward force. This causes a decrease in the number of molecules at the surface. The remaining molecules have to spread themselves to cover this area. Obviously they have to spread themselves, creating this attractive force that makes the surface like a stretched balloon. This force is of the typically of the order of 10^-10.
You can see now why a human cannot float on water because s/he is at least 12 orders more than the surface tension force.
When you apply a force on a hard object, there is not enough space between the molecules to push them closer beyond a point. They can only resist the change that is trying to push them which is the action force. In turn then exert a reaction force on the object stopping them quickly. Instead if the object is soft, then the molecules can get deformed somewhat like a seat belt trying to restrain you from stopping soon in a car when you suddenly break. This causes an increase in time of interaction.
The force you exert on the ground is the action and the reaction is the force the ground exerts on you on account of you pushing the ground.
You cannot expect to float on the air if you kick on it, but that's what birds are good at doing and here other physics principles come into play. It too bad Yang we're all much too fat for these forces to have any significant effect, unless of course you're capable of pushing a huge amount of air down like an helicopter!
10. To answer your first question, if an object is stopped from 180 ms-1 to rest in 5 seconds the acceleration is
a = 0 - 180/5 = 36 ms-2. If the mass of the object is 1 kg, then the force exerted by the earth is indeed 36 N. However if the speed is brought down in 5 minutes the time is now 60 x 5 = 300 seconds. The force would be much less because the acceleration is
a = 0 - 180/300 = 0.6 ms-2. The 1 kg mass would now experience a force of only 0.6 N!
We need something to balance our weight downward. The reaction of the ground balances our weight and we can therefore stand. If we were giants then the earth beneath our feet might give way. Alternately if we were bloated as balloons, then the upthrust on the balloon being greater than the weight, we could actually fly and float as a balloon!
The stones are there to absorb a good amount of the force and keep the 'sleepers' in place. Periodically the maintenance guys will pull them back together as the railway tracks are in use. A rigid framework would crack and not withstand such huge forces.
When you try to fill it at high speed you're not giving enough time for the air which was originally there to get out. It's like everyone trying to get into a crowded train/bus/elevator etc. Someone's got to be left behind!
11. i). What haven't you got yet? Yes the reaction force exerted by the ground it equal to the force the object exerted on the ground. I had given you numerical values earlier, 36 N and 0.6 N. I can't recall any other force here.
ii). Yes it is sometimes joked that if all the Chinese population jumped on to the ground at the same time, you'll have a major calamity in Europe and rest of the world. Of course you're not taking the elasticity of the ground here!
Indeed using the same reasoning, from Newton's third law, if you move forward you must be pushing the earth backward. If all of us, including from those down under in Australasia, then you would be exerting a force strong enough to influence the earth's orbital motion. In all this we're speaking of hypothetical situations and you must not get carried away Yang. They're good reasoning questions and help us think. I generally discuss such questions with my students including the transmission of power using transformers and so on.
12. You're right about wanting the force of gravity to be the same on the two cases, and indeed it is. The object has the same weight in the two cases. What is different is similar to what happens when say you're driving a car and apply the brakes suddenly. If you're wearing the seat belt (similar to your soft ground) then the time of interaction is increased and you don't experience a very large force on yourself. However, if you were not wearing the seat belt, then you're abruptly stopped by say the steering wheel (similar to your hard ground), assuming that you don't have air bags, the force is very large and you could be badly hurt. In both the cases it is the same person on the drivers seat. But the force on him/her is different. I hope this'd help clarify the matter a little.
13. I'll try to address this problem with a different lingo, if it'll help. Have you heard about the concept of impulse? If you jump onto a hard ground, two possibilites can result. In both cases the weight of the person is the same. However,
a) if you flex your knees, you don't get hurt because the time of interaction is increased (similar to soft ground!). This is because the impulsive force is less.
b) if you don't flex your knees, you can get badly hurt because the time of interaction is less(similar to hard ground). The impulsive force is larger in this case.
A similar situation exists in different ball sports as well and a martial art such as karate. When the time of impact is reduced, the hurt inflicted is more.
14. In my opinion, one of the best places where you can find out more on the physics of football would be to read Asai, Akatsuka and Haake’s article featured in the June 1998 issue of Physics World. You can read about the aerodynamics of ball sports, current research into football motion and more by clicking on the link below to read and savor.
http://physicsweb.org/article/world/11/6/8/1
0 = (10)(0.1) + (0.001)v
0 = 1 + 0.001v
v = 3.7 ms-1