LAWS OF MOTION LAWS OF MOTION MASSES CONNECTED

LAWS OF MOTION

LAWS OF MOTION MASSES CONNECTED BY SPRINGS/STRINGS

LAWS OF MOTION MASSES CONNECTED BY SPRINGS/STRINGS: i) When two bodies of masses m 1 and m 2 connected by a spring are pressed and released m 1 a 1 = m 2 a 2 , where a 1 and a 2 are their initial accelerations after release. ii) If a force F is applied on a body of mass m 1 which connected to another body of mass m 2 and the two bodies move with accelerations a 1 and a 2 respectively, then m 2 F = m 1 a 1 + m 2 a 2 m 1 F a 2 a 1

LAWS OF MOTION iii) A force F is applied on the massless pulley as shown in the figure, string connected to the block on smooth horizontal surface. Then T m T F = 2 T and T = mablock F

LAWS OF MOTION If the block moves a distance ‘x’ the pulley moves x/2 (total length of the string remains constant) Therefore acceleration of the pulley = ablock /2 = = =

LAWS OF MOTION MCQ S 1) In the following figure, the pulley is massless and frictionless. There is no friction between the body and the floor. The acceleration produced in the body when it is displaced through a certain distance with force ‘P’ will be T A M a) b) p T c) d)

LAWS OF MOTION APPARENT WEIGHT IN A LIFT OR ELEVATOR

LAWS OF MOTION Apparent weight in a lift or elevator: A man of mass ‘m’ is inside an elevator: a) Elevator accelerates up: Relative to earth N – mg = ma (M = mass of man) Apparent weight = N = m(g+a) Tension in the cable T = (Melevator +Mman)(g+a).

LAWS OF MOTION b) Elevator accelerates downward: Relative to earth mg - N = ma Apparent weight = N = m(g – a) Tension in the cable T = (Melevator +Mman)(g - a). Same effect is felt when the elevator goes up with retardation (a = -g) if an elevator falls freely (cable breaks) a = g hence N =0 i. e apparent weight of a body in a free fall = 0

LAWS OF MOTION c) Elevator moves up or down with uniform velocity: Relative to earth N – mg = 0 Apparent weight = N = mg Tension in the cable T = (Melevator + Mman)g

LAWS OF MOTION 1) A lift is going up with uniform velocity. When brakes are applied, it slows down. A person in that lift, experiences……. . a) more weight b) less weight c) normal weight d) zero weight MCQ S

LAWS OF MOTION 2) A man drops an apple in the lift. He finds that the apple remains stationary and does not fall. The lift is…. a) going down with constant speed b) going up with constant speed c) going down with constant acceleration d) going up with constant acceleration

LAWS OF MOTION MAN INSIDE AN ARTIFICIAL SATELLITE

LAWS OF MOTION Man inside an artificial satellite a) An artificial satellite orbiting the earth in a circular orbit is a freely falling body because its centripetal acceleration is equal to the acceleration due to gravity in that orbit. b) In the above case, apparent weight of the man inside the satellite is zero. mg - N (V 0 = orbital velocity)

LAWS OF MOTION But v 0 Hence N=0

LAWS OF MOTION MCQ S 1) The apparent weight of a man is zero in an artificial satellite orbiting the earth in a circular orbit is a freely falling body because ? a) Its centripetal acceleration is equal to g b) Its centripetal acceleration is not equal to g c) Its centripetal acceleration is greater than gequal to g d) Its centripetal acceleration is less than g

LAWS OF MOTION TENSION IN STRING SUPPORTING BOB

LAWS OF MOTION TENSION IN STRING SUPPORTING BOB: a) A bob of mass ‘m’ is suspended by a string. b) When the bob is stationary, the tension ‘T’ in the string is given by T = mg c) When the bob is pulled up with an acceleration ‘a’ the tension ‘T’ in the string is given by T=m(g +a). d) When the bob is allowed to move down with an acceleration ‘a’, the tension ‘T’ in the string is given by T = m(g – a).

LAWS OF MOTION e) When the bob is pulled up or moving down with uniform velocity ‘v’, the tension ‘T’ in the string is given by T = mg. f) When the bob is falling freely, the tension ‘T’ in the string becomes zero. g) When the suspended bob is moving horizontally with an acceleration ‘a’, the tension ‘T’ in the string is given by T=

LAWS OF MOTION h) A bird is in a wire cage hanging from a spring balance, when the bird starts flying in the cage, the reading of the balance decreases. i) In the above case, if the bird is in a closed cage or air – tight cage and it hovers in the cage, the reading of the spring balance does not change. j) In the above case for a closed cage if the bird accelerates upward reading of the balance is R = Wbird + ma, where m is the mass of the bird and a its acceleration.

LAWS OF MOTION 1)A bird is sitting in an air tight cage suspended from a spring balance. If the bird starts flying, then the reading of the balance will……… MCQ S a) becomes zero b) increases c) decreases d) remains unchanged

LAWS OF MOTION Thank you…