Edexcel PHYSICS Doubts

[elemis]

i'll post soon

Its been nearly an hour !

[nid404]

Guys I need help with this Physics question. Please assist!

The questions are attached.

1) I got velocity as ms^-1

2) a) I got velocity as ms^-1
    b) I got force as 100N

3) I explained that The boy goes to the right of the boat, and according to Newton's third law, the boat exerts an equal and opposite force     moving away from the boy.

4) Confused about the calculations involved.

5) Please help on this as well, I will check my answers when someone has posted.

1) 8/1000X 358= 91 X v
v=0.03m/s

2) a)5X4= 100Xv
 v=0.2m/s
b) F=ma
a=v/t
If you could tell me what you did?

3) that's right

4) astar has done this somewhere...I'll look for it

5) a) linear momentum of a body is the product of mass and velocity, and for a system equal to the vector sum of the products of mass and velocity of each particle in the system.

b) IIIrd statement is the second law
F=ma
F=m(v-u)/t
c) IInd

d) Not sure but i think he assumes it's a head-on collision.

e) should be using photogates. Will look for the entire setup.

[M-H]

Guys I need help with this Physics question. Please assist!

The questions are attached.

1) I got velocity as ms^-1

2) a) I got velocity as ms^-1
    b) I got force as 100N

3) I explained that The boy goes to the right of the boat, and according to Newton's third law, the boat exerts an equal and opposite force     moving away from the boy.

4) Confused about the calculations involved.

5) Please help on this as well, I will check my answers when someone has posted.

1st 2nd and 3rd are right..
4th and 5th, il b trying!! so sorry for the delay!..i know the diagram for the 4th one, but i don't know how to work it out cz we haven't yet started those stuff. anyway, ill try to get it asap!

[Saladin]

I only have one thing to add.

The hero must be pretty fat to be 100 KG....

[M-H]

I only have one thing to add.

The hero must be pretty fat to be 100 KG....

yeah fatso

[Saladin]

What do you guys get?

This is what I did:

Before collision

Horizontal


After Collision

Horizontal
Cue=

Black=

Therefore resolving horizontally to find the speed of the Cue ball





Vertical
Black=
Black=

At this point, it is safe to presume that as it is an elastic collision, they will move at equal and opposite angles. This means that the cue ball will also move at 45^o to the right.

Thus now finding the velocity:




Thus, this means that it will end up in the bottom right pocket as it is 45 to the bottom right, and therefore at the right speed, it will fall into the little pocket on the bottom right.

[Saladin]

This is another problem....


There are 6283 milliradians in a complete circle. The army’s rounding of this to 6400 mils causes an error.

   a   How far sideways from the target could this rounding cause an artillery shell to be when aimed at a target 20km away?

I know this is most probably so simple that I just cant see it infront of me. But I just can't seem to be able to solve this...

The answer to this by the way is 36.6 cm.

[Saladin]

This is another problem....


There are 6283 milliradians in a complete circle. The army’s rounding of this to 6400 mils causes an error.

   a   How far sideways from the target could this rounding cause an artillery shell to be when aimed at a target 20km away?

I know this is most probably so simple that I just cant see it infront of me. But I just can't seem to be able to solve this...

The answer to this by the way is 36.6 cm.

I already got this answered. Thanks everyone that had a look.

[Saladin]

Hey guys, can you please take a look at this for me?

I want to be sure about question 2.

You see there are two answers for part 2(a)

1. They act on the same body or do not act on different bodies
2. They are different types of, or they are not the same type of, force

I agree with the first one, but I believe that the second answer is incorrect, can anyone please tell me as to why the second answer is correct or if I am the one who is correct.

Because Newton's third law only talks about two separate forces on two separate bodies. The reaction force is a different kind of force in comparison to gravity....

Can anyone help?

[Saladin]

Hey guys, can you please take a look at this for me?

I want to be sure about question 2.

You see there are two answers for part 2(a)

1. They act on the same body or do not act on different bodies
2. They are different types of, or they are not the same type of, force

I agree with the first one, but I believe that the second answer is incorrect, can anyone please tell me as to why the second answer is correct or if I am the one who is correct.

Because Newton's third law only talks about two separate forces on two separate bodies. The reaction force is a different kind of force in comparison to gravity....

Can anyone help?

In fact, the whole question is wrong because in circular motion, there is acceleration! So, acceleration is caused by a resultant force!

How can this be?

[nid404]

In circular motion there is centripetal acceleration cause by the centripetal force acting towards the center of the rotating system.

As far as your question is concerned, lol, I don't quite get the first one itself  I feel dumb 

[nid404]

I'll try though.

[S.M.A.T]

Hey guys, can you please take a look at this for me?

I want to be sure about question 2.

You see there are two answers for part 2(a)

1. They act on the same body or do not act on different bodies
2. They are different types of, or they are not the same type of, force

I agree with the first one, but I believe that the second answer is incorrect, can anyone please tell me as to why the second answer is correct or if I am the one who is correct.

Because Newton's third law only talks about two separate forces on two separate bodies. The reaction force is a different kind of force in comparison to gravity....

Can anyone help?

Newton's third law,
'When a body 'A' exerts a force on body ''B',then body 'B' exerts a force of the same type on body 'A' which is equal in magnitude but opposite in direction.'
From the above statement it could be understand that
1)The two forces are equal in magnitude but opposite in direction
2)Two forces must be acting on two separate body.(in the above statement 'A' &'B'.)
3)The two forces must be same type of force.
4)The two forces must be acting for the same period of time i.e if body 'A' exerts a force for 't' second than on body 'B' then body 'B' exerts a force for 't' second than on body 'A'.

In the question,
1) 'F-1' &'F-2' are acting on the same body.So it is violating the point 2.
2) 'F-1' &'F-2' are not same type of force.'F-1' is normal reaction force whereas 'F-2' is weight.So it is violating point 3.

[Saladin]

Newton's third law,
'When a body 'A' exerts a force on body ''B',then body 'B' exerts a force of the same type on body 'A' which is equal in magnitude but opposite in direction.'
From the above statement it could be understand that
1)The two forces are equal in magnitude but opposite in direction
2)Two forces must be acting on two separate body.(in the above statement 'A' &'B'.)
3)The two forces must be same type of force.
4)The two forces must be acting for the same period of time i.e if body 'A' exerts a force for 't' second than on body 'B' then body 'B' exerts a force for 't' second than on body 'A'.

In the question,
1) 'F-1' &'F-2' are acting on the same body.So it is violating the point 2.
2) 'F-1' &'F-2' are not same type of force.'F-1' is normal reaction force whereas 'F-2' is weight.So it is violating point 3.

Thanks man!

[Saladin]

Just to clarify, the two are different forces because one is not a cause of the other...