Question

A 56 kg boat that is 7.6 m in length is initially 7.6 m from the pier. A 39 kg child stands at the end of the boat closest to the pier. The child then notices a turtle on a rock at the far end of the boat and proceeds to walk to the far end of the boat to observe the turtle. 7.6 m 7.6 m How far is the child from the pier when she reaches the far end of the boat?

Answer 1

Answer:

Explanation:

Given

mass of boat $M=56\ kg$

mass of Person $m=39\ kg$

Length of boat $L=7.6\ m$

Center of mass of boat and child

$x_{cm}=\frac{M\times 0.5L+m\times L}{M+m}$

$x_{cm}=\frac{56\times 0.5\times 7.6+39\times 7.6}{56+39}$

$x_{cm}=\frac{509.2}{95}$

$x_{cm}=5.36---1$

when child reaches another end of boat, boat is shifted by x cm

Center of mass of the system will remain the same

$x'_{cm}=\frac{M\times (x+0.5L)+m\times x}{M+m}$

$x'_{cm}=\frac{56\times (x+0.5\times 7.6)+39\times x}{56+39}----2$      

Equating 1 and 2 we get

$509.2=95x+212.8$

$x=3.12\ m$  

Answer 2

The child is 12.08 m away from the pier.

Initially let the center of mass of the boat be at x distance from the far end. When the child moves toward the far boat shifts to maintain the position of the center of mass.

given that the mass of boat M = 56kg

                       mass of the child = 39kg

                       distance of boat from near end to the pier = 7.6 m

                        length of boat =7.6m

the center of mass in the initial condition, taking far end as the origin

$x =\frac{56*(7.6/2)+39*7.6}{56+39}\\\\x = 5.36 m$

now let the boat moves a distance X from the initial position of the far end as the child moves toward the far end

$x =\frac{ 56*(x+7.6/2)+39+x}{56+39}\\\\X = 3.12 m$

So now the child is

7.6 + 7.6 - 3.12 = 12.08 m away from the pier

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