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(s) related to particle $m$
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(s) related to particle $m$
- A
Its kinetic energy is $K_f = \left( {\frac{{mM}}{{m + M}}} \right)gh$
- B
$v_1 = v_0 \left( {\frac{{M - m}}{{M + m}}} \right)$
- C
The ratio of its final kinetic energy to its initial kinetic energy is $\frac{{{K_{\text{f}}}}}{{{K_i}}} = {\left( {\frac{M}{{m + M}}} \right)^2}$
- D
It moves opposite to its initial direction of motion
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 $h$. 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 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 $kg-m/sec$ )
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A man is standing on a cart of mass double the mass of man. Initially cart is at rest. Now man jumps horizontally with relative velocity $'u'$ with respect to cart. Then work done by internal forces of the man during the process of jumping will be :
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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]