Thursday, September 27, 2007

Two Questions (MCQ) on Momentum Conservation

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 v1 and v2 are the average velocities of the skaters respectively, we have 40 v1 + 50 v2 = 0.

Therefore, 40 v1 = – 50 v2.

Since v1 = s1/t and v2 = s2/t where s1 and s2 are the displacements, we have

40 s1 = –50 s2

Considering the magnitudes of displacements, 40 s1 = 50 s2.

Since s1+ s2 (which is the total distance covered by the two scaters) is 4 m, s2 = 4–s1 so that

40 s1 = 50(4–s1) from which s1 = 2.22 m and s2 = 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 m1s1 = m2s2 where m1 and m2 are the masses of the spheres and s1 and s2 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, s1+s2 = 12R so that m1s1 = m2(12R–s1).

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

R3s1 = (2R)3(12R–s1) from which s1 = 10.67 R.

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