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URGENT!!!! PHYSICS MCQ QUESTIONS

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Tammet:
Anybody there???????????

Tammet:
Anybody there??

~Mimi~:
Nov 2009
 
#14


the easiest method seemed to be substitution here.
assume m=2, Initial v= 4,
so E(initial)= 1/2 *mv^2= 16 joules

now at the top, the vertical component of velocity is 0. There is no acceleration in the horizontal direction, so the horizontal component of velocity (the only velocity at the top) is the same as the beginning.
This is  v cos 45, again using v=4, horizontal component of velocity at the top= 4 cos 45
now find E at the top
E(new)= 1/2 * 2 * (4 cos 45)^2 = 8 joules
now compare the new E to the initial E.  8/16=1/2=====> 0.5E (The answer is A)....

astarmathsandphysics:
13. Tension=torque at Q/r=3/0.1=30
Torque at P=tension*distance=30*0.15
14. horizontal component of velocity=Vcos45
at top only horizontal component is only one since vertical velocity is zero so KE is 1/2m(vcos45)^2 =1/4mv^2 =1/2of inital KE
15. From conservation of momentum 2*2=4*1 so velocity of 1kg mass is 4
KE=1/2mv^2 for both trolleys 1/2*2*2^2+1/2*1*4^2=12
29 A use F=EQ

astarmathsandphysics:
nd q 6 nd 9 nov 2009
6 B the ball is bouncing. They have it upside down
9 A separation speed=approach speed=2u

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