PHYSICS (PHY 101)​​ 

QUESTION​​ 1

From Newton’s law of gravitation, the force of attraction between a body of mass​​ m1​​ and another body of mass​​ m2, separated by a distance​​ r​​ is given by the equation:

F=GM1M2r2

Find the dimension for the gravitational constant​​ G.​​ 

ANSWER​​ -

F=GM1M2r2

​​ Fr2=GM1M2

G=Fr2M1M2

dimensionally→ F=kgms-2=MLT-2,    r2=m2=L2,    M1=kg=M,     M2=kg=M

G=MLT-2∙L2M∙M=ML3T-2M2=M∙M-2L3T-2=M-1L3T-2

∴G=M-1L3T-2

QUESTION​​ 2

What is the resultant force when a 25N force acting to the north and another 55N force acting to the south are applied on an object?

ANSWER

 25N

 ​​ 55N

resultant force=Fsouth-Fnorth          since they are in opposite directions 

resultant force=55N-25N=30N

r=30N  in the direction of the larger force, i.e. towards the south direction

QUESTION 3

For what values of a are​​ A→=ai-2j+k​​ and​​ B→=2ai+aj-4k​​ perpendicular?

ANSWER

for perpendicular vectors, θ=90°  ;       but cos⁡90°=0

∴A→∙B→=0  ;      ai-2j+k∙2ai+aj-4k=0

→      2a2-2a-4=0

solving quadratically,  2a2-4a+2a-4=0

→    2aa-2+2a-2=0

→    2a+2a-2=0

→    2a+2=0       or       a-2=0

→    2a=-2           or        a=2

→     a=-22=-1  or       a=2

∴      a=-1 or 2

QUESTION 4​​ 

Find the maximum speed with which an unpowered car of mass 1200kg can take at a corner of radius 15m, if the coefficient of friction between the tyres and the road is​​ 0.4

ANSWER

F=μR

ma=μmg

a=μg

Fc=mv2r  ;         ma=mv2r

v2=ar  ,        but,  a=μg

v2=μgr

QUESTION 5

A 30kg block is suspended from a rope and lowered from a height of 15m to a height of 3m, experiencing a downward acceleration of 2ms-2. Find the work done on the block by the rope.

ANSWER

m=30kg,        h2=15m,  h1=3m,       g=9.81ms-2

→          workdone=mgh=mgh2-h1

→          workdone=30×9.81×15-3=30×9.81×12=3528

∴           workdone=3528joules