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