Question

A wedge (free to move) of mass 'M' has one face making an angle a with
horizontal and is resting on a smooth rigid floor. A particle of mass 'm' hits the Vo
inclined face of the wedge with a horizontal velocity vo. It is observed that the
particle rebounds in vertical direction after impact. Neglect friction between
a M
particle and the wedge & take M = 2m, Vo=10m/s, tana=2, g = 10m/s2

Assume that the inclined face of the wedge is sufficiently long so that the particle hits the same fac
once more during its downward motion. Calculate the time elapsed between the two impacts.​

Answer 1

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put (-4.7,-9.6){\line(1,0){20}}\put (15.3,-9.6){\line(0,1){40}}\put (-4.7,-9.6){\line(1,2){20}}\multiput (-20,17.5)(27,0){2}{\circle{3}}\put (-18.5,17.5){\vector(1,2){2}}\put(7,19){\vector (1,2){4}}\put (-18.5,17.5){\vector(1,0){8}}\put (-18.5,17.5){\vector(2,-1){6}}\put(7,19){\vector (-2,1){4}}\put(7,19){\vector(0,1){8}}\put(7,40){\circle {3}}\put(-21.6,20.4){ m }\put(-10.3,17.4){\tiny\vec{v_0}}\put(5.5,27.5){\tiny\vec{v'}}\put(16.5,-9.6){ M }\multiput(-17,18.7)(22.3,1.8){2}{\tinya}\put(-2.6,-8.5){ a }\put(15.3,10){\vector (1,0){10}}\put(19,5.5){ V }\end{picture}$

Let $v'$ and $V$ be the velocities attained by the particle and the wedge respectively after collision.

When we apply linear momentum conservation in the direction parallel to the inclined edge, we get,

$mv_0\cos a=mv'\sin a+MV\cos a$

Since $M=2m,$

$v_0\cos a=v'\sin a+2V\cos a\\\\\\v'=(v_0-2V)\cot a\quad\quad\dots (1)$

When we apply linear momentum conservation in the direction perpendicular to the inclined edge, we get,

$mv_0\sin a=MV\sin a-mv'\cos a\\\\\\v_0\sin a=2V\sin a-v'\cos a\\\\\\v'=(2V-v_0)\tan a\quad\quad\dots (2)$

From (1) and (2), we get,

$v_0=2V$

Then from (2), we get that,

$v'=0\ m\ s^{-1}$

It means the particle doesn't go upward after collision! Hence the two impacts occur simultaneously so the time taken between them is 0 seconds.