Question

the center of mass of two masses m and m' moves by distance x/5 when mass m is moved by distance x and m' is kept fixed. the ratio m'/m is​

Answer 1

Answer:

option 2)...4

Explanation:

x/5=MX/m+m'

m'+m=5m

m'/m=4

Answer 2

Answer:

The correct option is option 2) 4

Explanation:

Here we have been given that the center of mass of two masses m and m' has been moved by a distance of $\frac{x}{5}$ when the mass m has been moved by a distance of x and the other mass m' is fixed at the place.

We have to find the ratio of m' and m.

We know that the formula to find the position of the center of mass is as below;

Let us assume that R is the previous center of mass of the two given masses and so the position of the new center of mass is R + $\frac{x}{5}$

So the formula for the center of mass is:

$R = \frac{m_{1}r_{1}+m_{2}r_{2}}{m_{1}+m_{2}}$

Here we have,

$m_{1} = m$

$m_{2}= m'$

$r_{1}$ = position of mass m

$r_{2}$ = position of mass m'

$R= \frac{mr_{1}+m'r_{2}}{m+m'}$     (equation 1)

Now as per the conditions the mass m is moved by a distance x and m' is fixed so we have,

$r_{3} = r_{1}+x$

$r_{2}$ is same.

Now we have,

$R + \frac{x}{5} = \frac{m(r_{1}+x)+m'r_{2}}{m+m'}$

⇒ $R + \frac{x}{5} = \frac{(mr_{1}+mx)+m'r_{2}}{m+m'}$

⇒ $R + \frac{x}{5} = \frac{mr_{1}+m'r_{2}+ mx}{m+m'}$

⇒ $R + \frac{x}{5} = \frac{mr_{1}+m'r_{2}}{m+m'} + \frac{mx}{m + m'}$

⇒  $R + \frac{x}{5} = R + \frac{mx}{m + m'}$

⇒ $R + \frac{x}{5} - R = \frac{mx}{m + m'}$

⇒ $\frac{x}{5} =\frac{mx}{m + m'}$

⇒ $\frac{1}{5} =\frac{m}{m + m'}$

⇒ m + m' = 5m

⇒ m' = 5m - m

⇒ m' = 4m

⇒ $\frac{m'}{m} = 4$

Therefore the ratio  $\frac{m'}{m}$ is 4.

Hence option 2) 4 is correct.