Two sphere A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v along the x -axis. After the collision, B has a velocity v/2 in a direction perpendicular to the original direction. The mass A moves after collision in the direction
same as that of B
opposite to that of B
C.
Here, Pi = m2vi +m2 x 0
A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is wound around the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 rev/s2 is,
25 N
50 N
78.5 N
78.5 N
D.
78.5 N
Radius, r = 0.5 m
Angular acceleration,
Torque produced by the tension in the string,
= T x R = T x 0.5 = T/2 Nm ... (i)
We know,
T = 1α ... (2)
From equations (i) and (ii)
Therefore,
The ratio of the accelerations for a solid sphere (mass m and radius R) rolling down an incline of angle without slipping and slipping down the incline without rolling is,
5:7
2:3
2:5
2:5
A.
5:7
A solid sphere rolling without slipping down an inclined plane is,
The upper half of an inclined plane of the inclination is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by,
C.
mg sin . L = mg cos