Two spheres of masses m1 and m2 (m1>m2) respectively are tied to the ends of a light, inextensible string which passes over a light frictionless pulley. When the masses are released from their initial state of rest, the acceleration of their centre of mass is:

Q. Relation between

m_{1}

and

m_{2}

- 4 m_{1} - 4 m_{2} = 3 m_{1} - 6 m_{2}

Three different objects of mass $m_{1}$ , $m_{2}$ and $m_{3}$ are allowed to fall from rest and from the same point along three different frictionless paths. The speed of the three objects on reaching the ground will be in the ratio of

Q. An object of mass

m_{1}

collides elastically with another object of mass

m_{2}

, which is at rest. After the collision, the objects move with equal speeds in opposite directions. The ratio of the masses,

m_{2} : m_{1}

is -

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The mass of object X is M1 and that of object Y is M2. Keeping their kinetic energy constant, if the velocity of object Y is doubled the velocity of object X, what will be the relation between their masses?

The correct option is B M1=4M2Mass of object X = M1 , Velocity of object X = v1Mass of object Y = M2, Velocity of object Y = v2Also, v2 = 2v1K.E is constant.K.E1 = K.E212 M1 v21 = 12 M2 v2212 M1 v21 = 12 M2 (2v1)212 M1 v21 = 12 M2 4 v21M1 = 4 M2

The mass of object X is $M_{1}$ and that of object Y is $M_{2}$ . Keeping their kinetic energy constant, if the velocity of object Y is doubled the velocity of object X, what will be the relation between their masses?