Question

A small wedge of mass M slides on the bigger wedge
initially resting on the smooth horizontal ground as shown in
the figure. The mass of the bigger wedge is 6M. When the
smaller wedge reaches the base of the bigger wedge, the
distance moved by the bigger wedge is
KAL
(B) 4015
C) 2L/3
(D) 3L17
6M​

Answer 1

When the  smaller wedge reaches the base of the bigger wedge, the  distance moved by the bigger wedge is 3L/7

Explanation:

As we know that there is no external force on the system of two wedge along horizontal direction

So we will have

$F_{net} = 0$ along x direction

so we can say that

$\Delta x_{cm} = 0$

so we can say

$m_1x_1 + m_2x_2 = 0$

now we have

$M(3L - x) + (6M)(-x) = 0$

$3ML = 7 Mx$

$x = \frac{3L}{7}$

#Learn

Topic : Motion of center of mass

https://brainly.in/question/4566679

Answer 2

Answer:

Explanation:

by system

summation fx=0

acceleration with respect to ground

let acceleration of mass 6M be A

and acceleration of mass M be a wrt to wedge

acceleration of mass M wrt to ground = acostheta-A in x axis

acceleration of mass 6M wrt to ground = -A

summation Fx=

6M(-A)+M(acostheta-A) = 0

-7A = -acostheta

7A=acostheta

let displacement of 6M=X

and displacement of M = x

7X = xcostheta

X=4L / 7

X=4L/7