The upper half of an inclined plane with inclination $\phi$ is perfectly smooth, while the lower half is rough. $A$ body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is given by-
The upper half of an inclined plane with inclination $\phi$ is perfectly smooth, while the lower half is rough. $A$ body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is given by-
- A
$2 \sin \phi$
- B
$2 \cos \phi$
- C
$2 \tan \phi$
- D
$ \tan \phi$
Similar Questions
A particle of mass $m$ moving horizontally with $v_0$ strikes $a$ smooth wedge of mass $M$, as shown in figure. After collision, the ball starts moving up the inclined face of the wedge and rises to $a$ height $h$. The final velocity of the wedge $v_2$ is
A particle of mass $m$ moving horizontally with $v_0$ strikes $a$ smooth wedge of mass $M$, as shown in figure. After collision, the ball starts moving up the inclined face of the wedge and rises to $a$ height $h$. The final velocity of the wedge $v_2$ is
A block of mass $M$ has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at $x=0$, in a co-ordinate system fixed to the table. A point mass $m$ is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is $\mathrm{x}$ and the velocity is $\mathrm{v}$. At that instant, which of the following options is/are correct?
(image)
$[A]$ The $x$ component of displacement of the center of mass of the block $M$ is : $-\frac{m R}{M+m}$.
[$B$] The position of the point mass is : $x=-\sqrt{2} \frac{\mathrm{mR}}{\mathrm{M}+\mathrm{m}}$.
[$C$] The velocity of the point mass $m$ is : $v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}}$.
[$D$] The velocity of the block $M$ is: $V=-\frac{m}{M} \sqrt{2 g R}$.
A block of mass $M$ has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at $x=0$, in a co-ordinate system fixed to the table. A point mass $m$ is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is $\mathrm{x}$ and the velocity is $\mathrm{v}$. At that instant, which of the following options is/are correct?
(image)
$[A]$ The $x$ component of displacement of the center of mass of the block $M$ is : $-\frac{m R}{M+m}$.
[$B$] The position of the point mass is : $x=-\sqrt{2} \frac{\mathrm{mR}}{\mathrm{M}+\mathrm{m}}$.
[$C$] The velocity of the point mass $m$ is : $v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}}$.
[$D$] The velocity of the block $M$ is: $V=-\frac{m}{M} \sqrt{2 g R}$.
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