Question

Three men a,b and c of masses 40 kg,50 kg and 60 kg are standing on a plank of mass 90 kg ,which is kept on a smooth horizonatal plane . If a and c exchange their positions then mass b will shift

Answer 1

Answer:

1/3 m towards the right.

Step by step explanation:

The center of mass (CM) cannot shift position. Therefore, the CM of the system must remain the same before and after the exchange.

The equation of the position of the CM is given by:

$x_{cm}=\frac{m_Ax_A+m_Bx_B+m_Cx_C}{m_A+m_B+m_C+M}$

Before the exchange: $x_{cm1}=\frac{40(-2)+50(0)+60(2)}{40+50+60+90}=1/6\:m$

After the exchange: $x_{cm2}=\frac{60(-2)+50(0)+40(2)}{40+50+60+90}=-1/6\:m$

Therefore, mass B will shift by: $x_{cm1}-x_{cm2}=\frac{1}{3}\:m$ towards the right.

Answer 2

Centre of mass will remain conserved. .So mass B will shift 1/3 m towards right

Explanation:

Forces on the body are internal forces therefore, centre of Mass shall remain same before and after the exchange of weights.  

The formula of Centre of Mass Xcm= (MaXa+MbXb+McXc)/(Ma+Mb+Mc+M)

Where, Ma, Mb, Mc & M are mass of men a, b, c and plank respectively

Xcm= centre of mass

Xa,Xb & Xc are distance of centre of mass from men a, b & c respectively

CM before the exchange= {40x(-2)+50x(0)+60x2}/(40+50+60+90)

=1/6 m

CM after the exchange ={60x(-2)+50x(0)+40x2}/(40+50+60+90)

=-1/6 m

CM shifted by =(1/6)-(-1/6)=1/3 m toward right