Where Will They Meet?

Since the system is isolated so no other force except gravitational force acts between them. This implies the center of mass of the system remains at rest and for this to be true the bodies must meet at the center of mass of the system.

So the required distance is the position of center of mass w.r.t $7\,kg$ mass i.e. $\textbf{r}=\displaystyle\frac{3\times10+0\times7}{3+7}=\boxed{3\,m}$ .

For this problem, you could probably go out of your way doing integrals and doing tons of complicated stuff; all you need, however, is some of Newton's laws and a bit of physical intuition :).

Everything gets much simpler if we remember Newton's third law- equal and opposite forces. The two point masses will always exert equal forces on each other; the only difference will be the resulting accelerations. So it doesn't really matter that they're getting closer to each other and the gravitational attractions are increasing, because they're increasing equally for both masses. The one difference is the actual mass; remembering Newton's third: the 3 kg mass will have an acceleration of F/3 (F is changing but we don't care), while the 7 kg mass will have an acceleration of F/7. The acceleration of the 7 kg mass is 3/7 that of the 3 kg mass. Distance covered is directly proportional to acceleration, so the distance covered by the 7 kg mass is 3/7 that of the distance covered by the 3 kg mass! Let's call the distance the 3 kg mass goes x. Then 10=x+3/7•x=x(1+3/7)=x(10/7). So, x=7 (the distance covered by the 3 kg mass is 7 meters). Now we can either subtract that from the total distance, or multiply it by 3/7; it doesn't matter. Both operations give us the same result, the distance covered by the 7 kg mass (the point at which they meet starting from the 7 kg mass): 3 meters! Pretty great, right?

The two objects will meet at their center of mass. Let $x = 0 \, \text{m}$ be the position of the $7$ -kg object and let $x = 10 \, \text{m}$ be the position of the $3$ -kg object. Then,

$x_{\text{COM}} = \dfrac {(0 \, \text{m}) \cdot (7 \, \text{kg}) + (10 \, \text{m}) \cdot (3 \, \text{kg})}{3 \, \text{kg} + 7 \, \text{kg}} = 3 \, \text{m},$

so the displacement we seek is $3 \, \text{m} - 0 \, \text{m} = \boxed {3} \, \text{m}$ .

Since no external force is working except their mutual force, thus, dx(COM)= 0

So, mdx = Mdy (m = mass of smaller body, M = mass of larger body, dx = distance moved by the smaller body, dy = distance moved by the latger body)

Let the point where they meet be a metres from 7kg mass.

So, 3 X (10 - a) = 7 X a

a = 3

So, 3 metres is the answer.

Assume m=7kg & m"=3kg Force act on m & m" is 1.40 10^-11( by using the equation F=mm" G/r^2) Again,F=ma, a=F/m.By using this equation, I found the acceleration for m & m" is a= 2 10^-12 & a"= 4.67 10^-12 s=ut+.5at^2 s=at^2 t^2=s/a Assume the path passed by m is s & m" is (10-s). Here the duration of passing is same. So, 2s/a=2(10-s)/a" From this equation I found s is 2.99 m=3 m

They will of course meet at their centre of mass

The question would have been more challenging if it would have been WHEN will they meet!!! instead of Where will they meet!!!

as no other force than gravitation is acting the will meet at center of mass which is at 3 metres from 7. if we consider coordinate of 7kg to be 0 and that of 3kg to be 10 com=(3 10+7 0) \7+3=30\10=3 :)

this questin is similar to problems we do electrostatics just use the formula of gravitational force take distance 3 kg mass be X then from from 7 kg it will be 10-X and equate the forces.

as no external force acts on the composite system, therefore the center of mass of the system will be at rest or we can say that the two masses will meet at their center of mass. so in reference frame of 7 kg mass, the center of mass is at 3*10/3+7=3m

The two bodies are at rest (initially). We will consider the two bodies as one system. They will meet at their center of mass (center of mass of the two body system). Since there is only motion along one axis (say X). CM at X-axis is:

(m1.x1 + m2.x2)/(m1 + m2) [This is how you find Center of Mass of two body system like the one in question]

Here m1=3kg, x1=0m, m2=7kg, x2=10m. Thus Center of mass is at 7m from 3kg, or 7m from 7kg.

We take m1 to be the starting point(datum point), thus x1=0m.

first of all lets calculate the acceleration on 7kg mass using the standard formula for gravitaional force i.e. F= G $M_{1}$ $M_{2}$ / $r^{2}$ . Now the acceleration on 7kg is F/7 = 3G/100.

Let 'x' be the distance from 3kg mass where the particles meet. Therefore 10-x is the point from 7kg where the points meet. Using the formula x=1/2 X $a_{1}$ X $t^{2}$ and 10-x = 1/2 X $a_{2}$ X $t^{2}$ . Since the time of meeting is same for both the particles. On solving we get $t^{2}$ = 200/G On plugging this value in 10-x equation, we get the desired result i.e. Ans = $\boxed{3}$

In the absence of any external force, the center of mass will remain at rest as it was at the time of leaving the bodies. So, effectively we have to calculate the center of mass position from body of mass 7 kg.

When the two bodies collide they shall be at their center of mass and hence, the ans is 3 as the center of mass of the two bodies is 3 m from the larger mass.

Since this system only evolves through internal forces, the center of mass of the system remains constant. Therefore, they will meet at the center of mass of the system.

$Mr_{cm} = m_{1}r_{1} + m_{2}r_{2}$

Where $M$ is the total mass, $m_{1}$ is 3 kg, $m_{2}$ is 7 kg, $r_{1}$ is the distance the 7-kg block is from the 3-kg block, and $r_{2}$ is the distance the 3-kg block is from the 3 kg block.

When you evaluate this, $r_{cm}$ is 7 meters. The place where the two blocks meet is therefore $\boxed{3 meters}$ away from the 7-kg block.

Here the only forces in play are the mutual attraction forces between the two bodies which are action-reaction pair.As no external force acts on this two body system so the position of their center of mass will remain unchanged.So the only point at which they can meet is the center of mass of this system. Now suppose that the 3 kilo mass is at the origin and the 7 kilo mass is at the point (10,0).Then the center of mass lies on the x axis and we calculate it this way- x coordinate of the center of mass =(3 times 0 + 7 times 10)/(3+7)=7 SO the distance of the center of mass from the 3 kilo mass is 7.Hence the distance of the center of mass from the 7 kilo mass is (10 - 7) = 3.