Often you will find questions based on the law of conservation of momentum. Here is a question which may be interesting to you:

**Two skaters of masses 40 kg and 50 kg respectively are 4 m apart and are facing each other, holding the ends of a light rope. They advance by pulling on the rope, thereby reducing the separation between them. The approximate distances covered by them by the time their separation has reduced to zero are respectively**

**(a) 2.22 m and 1.78 m (b) 2 m and 2 m (c) 1.62 m and 2.38 m**

**(d) 4 m and zero (e) zero and 4 m **

The initial momentum of the system containing the two skaters is zero. The pulling forces they exert are internal forces (within the system) and hence the total momentum of the system must be zero throughout their motion. If v_{1} and v_{2} are the average velocities of the skaters respectively, we have 40 v_{1} + 50 v_{2} = 0.

Therefore, 40 v_{1} = – 50 v_{2}.

Since v_{1} = s_{1}/t and v_{2} = s_{2}/t where s_{1} and s_{2} are the displacements, we have

40 s_{1 }= –50 s_{2}

Considering the magnitudes of displacements, 40 s_{1} = 50 s_{2.}

Since s_{1}+ s_{2 } (which is the total *distance* covered by the two scaters) is 4 m, s_{2} = 4–s_{1} so that

40 s_{1} = 50(4–s_{1}) from which s_{1} = 2.22 m and s_{2} = 1.78 m [Option (a)]

The following MCQ is a simple one. But be careful not to be distracted to arrive at a wrong answer!

**Two spheres of the same material have radii R and 2R. They are released in free space with initial separation between their centres equal to 15R. If the only force between the spheres is the gravitational pull between them, the distance covered by the smaller sphere before collision is approximately**

**(a) 6.67R (b) 10.67R (c) 13.33R (d) ) 2.33R (e) 4.33R **

As in the previous question, the forces acting are internal forces and hence the total momentum of the system remains unchanged. The initial momentum of the system containing the two spheres is zero and hence the total momentum of the system must be zero throughout their motion.

Proceeding as before, we have m_{1}s_{1} = m_{2}s_{2} where m_{1} and m_{2} are the masses of the spheres and s_{1} and s_{2} are the distance covered by them *before collision*. But you have to be careful to note that when the spheres collide, the distance between their centres is (R+ 2R) = 3R so that the total distance covered by the spheres is 15R – 3R = 12R.

Therefore, s_{1}+s_{2}^{ }= 12R so that m_{1}s_{1} = m_{2}(12R–s_{1}).

Since the spheres are of the same material, their masses are directly proportional to their volumes which are proportional to the cubes of their radii. Therefore we have

R^{3}s_{1} = (2R)^{3}(12R–s_{1}) from which s_{1} = 10.67 R.