help needed!! 9702_w10_qp 11

[XPFMember]

Assalamoalaikum!!

I need help with Q:7 of Nov:2010 Paper 11
i thought it's D but mark scheme says it's B ???

 

[XPFMember]

And also Q:12 plz its urgent..have an exam tomorrow!

 

[Tweety-Angie]

yeah d answer of Q.7 is D, I don't know how come it be B in mark scheme..
and regarding Q.12 I think it is C.

 

[XPFMember]

y do u think so??btw ms says it's B

 

[Tweety-Angie]

the ans is D, cuz, velocity in vertical direction increases due to acceleration due to gravity, whereas the velocity in horizontal direction remains constant.

 

[XPFMember]

i was talkng abt Q:12...btw for Q:7 i also think the same

 

[Tweety-Angie]

hm..okei..all d best for ur test

 

[XPFMember]

thanks...btw it's full P1 and P2 wexam!!

 

[hassam]

q7....its D bcos air resistance will have effect on both components of motion and horizontal velocity will thus decreaSE
Q12...its b.jxt remember that for elastic collision relative speed of approach is equal to relative speed of separation after the collision and for inelastic collision relative speed of separation is zero

 

[XPFMember]

Assalamoalaikum!!

@hassam Canu plz elaborate a bit abt elastic and inelastic collisions?

 

[naavo.1234]

Important notes for elastic and inelastic collision

For a PERFECTLY elastic collision, THE RELATIVE SPEED OF APPROACH IS EQUAL TO THE RELATIVE SPEED OF SEPERATION.

If TWO identical objects (means having same MASS) approaches each other and collides, they exchange their velocities.

also for a perfectlly elastic collision KINETIC ENERGY IS ALSO CONSERVED BEFORE AND AFTER COLLISION. their are some cases where kinetic energy is not conserved that means some energy is transformed into heat or something else. But momentum is always conserved.

to find total momentum in closed or isolated system before collision of two object approaching each other:
m1u1+m2u2 = m1v1+m2v2 (THIS IS USED IF MASSES ARE NOT EQUAL)

u= initial speed for the two objects approaching
v= final speed for the two objects moving away from each other or moving in the same direction.

WARNING

MOMENTUM IS A VECTOR QUANTITY HAVING DIRECTION AND MAGNITUDE. IF YOU TAKE RIGHT SIDE TO BE POSITIVE VELOCITY THAN TAKE LEFT SIDE AS A NEGATIVE VELOCITY.

I HOPE I HAVE MADE MY SELF CLEAR. IF U NEED MORE HELP DON'T HESITATE TO ASK.

 

[XPFMember]

JAzak Allah Khair...May Allah Bless you!!

btw i have a doubt...relative speed of approach means...their initial velocities?? and so if always their vel. gets exchanged...that means..in the question in the first post.....then the object at rest will start moving and the object moving will cum to rest??? wont they stick together and move??

 

[naavo.1234]

THAT DEPENDS ON THE QUESTION!!

IF THE QUESTION SAYS THAT "ONE OBJECT OF MASS 2KG IS MOVING WITH 5m/s TOWARDS ANOTHER OBJECT OF MASS 2KG WHICH IS STATIONARY."

THE ANSWER IS THAT The object WHICH IS STATIONARY WILL MOVE WITH 5m/s in the same direction of the approaching object and the moving object which was moving previously will remain at rest.

i also have a question regarding deformation of solids if u could help me please refer to my latest post.!!!!

 

[hassam]

u cn remember by this example.....e.g if two objects are moving towards each other with v m/s this means that they ll strike each other very quickly and so relative speed of approach is found by adding two velocities however if they are moving away with different velocities relative speed of separation will be found by subtracting the smaller v from larger v

 

[XPFMember]

and in the first case what will be the result... i mean which direction will they move......i guess they'll cum to rest :?:

 

[DragonCub]

I think to Q12, both momentum (p) and kinetic energy (k.e.) are to be considered.
Suggest the masses of X and Y are both M, velocity of X before collision is V
Due to momentum conservation, in Experiment 1, before collision, total p = MV + 0 = MV
So after the collision total p is still MV - Y is moving with velocity of V to the right.
Then consider the k.e. Before collision: total k.e. = 0.5MV^2 + 0 = 0.5MV^2
After collision: 0.5MV^2. The same, so total k.e. is conserved -------> elastic collision

In experiment 2, before collision total p = MV
After collision, the trolleys stick together, so in the whole system, total mass = 2M, and thus the velocity becomes V/2
Then the k.e.: Before = 0.5MV^2, After = 0.5 * 2M * (V/2)^2 = 0.5 * 2M * (V^2)/4 = 0.25MV^2
Total k.e. is not conserved -------> inelastic collision

The answer should be B.