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 $h$. Choose the correct statement related to the wedge $M$
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 $h$. Choose the correct statement related to the wedge $M$
- A
Its kinetic energy is $K_f \left( {\frac{{4{m^2}}}{{m + M}}} \right)gh$
- B
$v_2 = \left( {\frac{{2m}}{{m + M}}} \right){v_0}$
- C
Its gain in kinetic energy is $\Delta K =$ $\left( {\frac{{4mM}}{{{{(m + M)}^2}}}} \right)\left( {\frac{1}{2}mv_0^2} \right)$
- D
All of the above
Similar Questions
Write the principle of conservation of mechanical energy for non-conservative force.
Write the principle of conservation of mechanical energy for non-conservative force.
A trolley of mass $200\; kg$ moves with a uniform speed of $36\; km / h$ on a frictionless track. A child of mass $20\; kg$ runs on the trolley from one end to the other ( $10\; m$ away) with a speed of $4 \;m s ^{-1}$ relative to the trolley in a direction opposite to the its motion, and Jumps out of the trolley. What is the final speed of the trolley ? How much has the trolley moved from the time the child begins to run?
A trolley of mass $200\; kg$ moves with a uniform speed of $36\; km / h$ on a frictionless track. A child of mass $20\; kg$ runs on the trolley from one end to the other ( $10\; m$ away) with a speed of $4 \;m s ^{-1}$ relative to the trolley in a direction opposite to the its motion, and Jumps out of the trolley. What is the final speed of the trolley ? How much has the trolley moved from the time the child begins to run?
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: $V=-\frac{m}{M} \sqrt{2 g R}$.
- [IIT 2017]
A bomb of mass $12\,\,kg$ at rest explodes into two fragments of masses in the ratio $1 : 3.$ The $K.E.$ of the smaller fragment is $216\,\,J.$ The momentulm of heavier fragment is (in $kg-m/sec$ )
A bomb of mass $12\,\,kg$ at rest explodes into two fragments of masses in the ratio $1 : 3.$ The $K.E.$ of the smaller fragment is $216\,\,J.$ The momentulm of heavier fragment is (in $kg-m/sec$ )
A particle of mass $m$ with initial kinetics energy $K$ approaches the origin from $x =+\infty$. Assume that a conservative force acts on it and its potential energy $V ( x )$ is given by $V ( x )=\frac{ K }{\exp \left(3 x / x _0\right)+\exp \left(-3 x / x _0\right)}$ where, $x_0=1 m$. The speed of the particle at $x =0$ is
A particle of mass $m$ with initial kinetics energy $K$ approaches the origin from $x =+\infty$. Assume that a conservative force acts on it and its potential energy $V ( x )$ is given by $V ( x )=\frac{ K }{\exp \left(3 x / x _0\right)+\exp \left(-3 x / x _0\right)}$ where, $x_0=1 m$. The speed of the particle at $x =0$ is
- [KVPY 2021]