CBSE Class 11-science Answered
Let the 2 kg mass is moving in a circular motion on the curves surface as shown in figure.
Let R be the radius of circular motion.
Normal force N acting on 2 kg mass is resolved into two components , one in vertical direction i.e. ( N sin15 )
and other one in horizontal ( N cos15 ) as shown in figure
Vertical component is balanced by weight of 2 kg mass, hence we have
........................(1)
where m is mass of block and g is acceleration due to gravity.
Horizontal component of normal force gives centripetal force required for circular motion.
...............(2)
By dividing eqn.(2) by (1) and we get after simplification as
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If the second person of mass 80 kg is moving with acceleration 5 m/s2 then he is subjected to a force ( 80 × 5 ) = 400 N
If the second person applies 1000 N force on first person of mass 50 kg , then total force of 1400 N is acting on combined mass 130 kg of both the persons.
( It is assumed both the persons are in contact ).
Hence the acceleration of both the persons = (1400 / 130 ) m/s2 .= 10.8 m/s2.
If the first person is initially at rest , then he is moving with speed V which is a function of time
V = (10.8 × t ) m/s
Relative velocity of block that is moving in circular motion with respect to first person is
Relative velocity = v - V