Physics MCQs thread.

[hassam]

 

[xHazeMx]

is it A?

 

[Zishi]

C, I guess.

 

[Hateexams93]

Help plz

its A, small masses will obtain higher kinetic energies in order to make the total momentum equal to zero. m1u1 + m2u2 = m1v1 + m2v2

WHY NOT B THEN ?

 

[Zishi]

Help plz

its A, small masses will obtain higher kinetic energies in order to make the total momentum equal to zero. m1u1 + m2u2 = m1v1 + m2v2

It can be solved using a very simple way. For elastic collisions, speed of approach = speed of separation

Speed of approach = u--u = 2u
Speed of separation must also be 2u which is A, (5u/3) - (u/3) = 2u

 

[Nibz]

*THIS THREAD IS CHANGED TO 'STICKY' NOW*

 

[Hateexams93]

It can be solved using a very simple way. For elastic collisions, speed of approach = speed of separation

Speed of approach = u--u = 2u
Speed of separation must also be 2u which is A, (5u/3) - (u/3) = 2u

so u have to MINUS to find the speed of approach ?

 

[xHazeMx]

It can be solved using a very simple way. For elastic collisions, speed of approach = speed of separation

Speed of approach = u--u = 2u
Speed of separation must also be 2u which is A, (5u/3) - (u/3) = 2u

so u have to MINUS to find the speed of approach ?

yeah, because velocity is a vector and they are in opposite directions, so u will have to minus them

 

[Hateexams93]

i know about their directions , i was asking about speed of approach, so the formula will be : speed of approach= U1 -U2 ??????

 

[xHazeMx]

i know about their directions , i was asking about speed of approach, so the formula will be : speed of approach= U1 -U2 ??????

use the equation, m1u1 + m2u2 = m1v1 + m2v2 , considering the directions and u will get the answer

 

[Hateexams93]

can some1 show me the working please ?

 

[xHazeMx]

can some1 show me the working please ?

just confirm it first and then i will show the working if it is correct. Is it B ?

 

[Hateexams93]

no its D

 

[Zishi]

can some1 show me the working please ?

The point(midpoint) at which the wheel Q is being rotated acts as the pivot. Moment = Force x perpendicular distance

Distance = radius =0.05m, you have the moment for it, you can easily calculate the force. The force will remain same for wheel P, so you can also calculate moment about P, using force acting on it and its radius.

 

[xHazeMx]

no its D

oh, sorry i took the diameter instead of the radius :/. here are the calculations.
Tension x perpendicular distance (from the centre of the wheel) = Torque
T x 0.1/2 (radius) = 3
T = 3/ 0.05 = 60

same equation for P

tension x perpendicular distance = Torque
60 x 0.15/2 = Torque
Torque = 4.5

 

[Hateexams93]

A projectile is launched at 45° to the horizontal with initial kinetic energy E.
Assuming air resistance to be negligible, what will be the kinetic energy of the projectile when it
reaches its highest point?
A 0.50 E B 0.71 E C 0.87 E D E

 

[hassam]

its A...SEE AT v initial is V...SO at highest point velcity will be V COS45...RI8...NW U LL UNDRSTND

 

[hassam]

 

[Hateexams93]

its A...SEE AT v initial is V...SO at highest point velcity will be V COS45...RI8...NW U LL UNDRSTND

can u show the working ..i mean the drawing plz

 

[xHazeMx]

its A...SEE AT v initial is V...SO at highest point velcity will be V COS45...RI8...NW U LL UNDRSTND

can u show the working ..i mean the drawing plz

vertical velocity= u cos 45
horizontal velocity = u sin 45
so, initial K.E. : the vertical and horizontal components of kinetic energy = 1/2 m ( u cos 45 ) + 1/2 m ( u sin 45 ) ......... ( 1/2 m v^2 formula )

so, sin 45 = 0.71
....cos 45 = 0.71 ( check them in the calculator )

so they will give the same kinetic energy for both velocities ( horizontally and vertically )

so at the highest point, (theta) = 90 ( for the vertical velocity ) , horizontal velocity doesn't change
so final kinetic energy = 1/2 m ( u cos 90 ) + 1/2 m ( u sin 45 )..................... ( horizontal velocity is constant ) ( cos 90 is 0 )
final kinetic energy = 1/2 m ( u sin 45 ) + 0 only which is half of the initial kinetic energy.