Centre Of Mass Ques 2
- A wedge of mass $M=4 \mathrm{~m}$ lies on a frictionless plane. A particle of mass $m$ aproaches the wedge with speed $v$. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by
(2019 Main, 9 April II)
(a) $\frac{2 v^2}{7 g}$
(b) $\frac{v^2}{g}$
(c) $\frac{2 v^2}{5 g}$
(d) $\frac{v^2}{2 g}$
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Answer:
Correct Answer: 2.( c )
Solution:
Formula:
Motion of Centre of Mass and Conservation of Momentum:
Key Idea Since, the ground is frictionless, so whe the particle will collide and climb over the wedge, then the wedge will also move. Thus, by using conservation laws for momentum and energy, maximum height climbed by the particle can be calculated. Initial condition can be shown in the figure below As mass $m$ collides with wedge, let both wedge and mass move with speed $v^{\prime}$. Then, By applying linear momentum conservation, we have Initial momentum of the system $=$ Final momentum of the system
$ \begin{aligned} m v+0 & =(m+4 m) v^{\prime} \\ \Rightarrow \quad v^{\prime} & =\frac{v}{5} \end{aligned} $
Now, if $m$ rises upto height $h$ over wedge, then by applying conservation of mechanical energy, we have Initial energy of the system $=$ Final energy of the system
$ \begin{aligned} & \frac{1}{2} m v^2+0=\frac{1}{2} m v^{\prime 2}+m g h+\frac{1}{2}(4 m) v^2 \\ & m v^2=(m+4 m) v^{\prime 2}+2 m g h \\ & \Rightarrow \quad v^2=5 v^{\prime 2}+2 g h \\ & \Rightarrow \quad v^2=\frac{1}{5} v^2+2 g h \\ & \Rightarrow \quad \frac{4}{5} v^2=2 g h \Rightarrow h=\frac{2 v^2}{5 g} \end{aligned} $
BETA
