(696) Problem 16: Two sprinters leave the starting gate at the same time at the

Question

(696) Problem 16: Two sprinters leave the starting gate at the same time at the beginning of a straight track. The masses of the two sprinters are 52.5 kg and 63 kg. Randomized Variables m1-52.5 kg m2 = 63 kg -.. 25% Part (a) A few seconds later, the first sprinter is ahead of the second by a distance 4.3 m. How far ahead of the second sprinter is the center of mass of these two sprinters, in meters? - 25% Part (b) If the speeds of the sprinters are 52 ms and 3.5 m/s, respectively, how fast, in meters per second, is the center of mass moving? 25% Part (c) What is the momentum of the center of mass, in kilogram meters per second? -là 2596 Part (d) How is the momentum of the center of mass related to the total momentum of the sprinters?

Explanation / Answer

(a) As distance between sprinters = 4.3 m

Let coordinates such that the sprinter behind is at x = 0

Hence, sprinter 2 is at x = 4.3 m

Now, their coordinates of center of mass is =

R = (m1x1 + m2x2)/(m1 + m2)

R = (63 x 0 + 52.5 x 4.3)/(63 + 52.5) = 1.9545 m

Hence, the center of mass is 1.9545 m ahead of the sprinter which is behind and is called the second speinter in question.

(b) as total momemtum = Momemtum of the center of mass

m1v1 + m2v2 = (m1 + m2)v

52.5 x 5.2 + 63 x 3.5 = (52.5 + 63) v

Or

493.5 = 115.5 v

v = 4.272 m/s

This is the required velocity of the center of mass.

(c) momentum of the center of mass is

P = MV = 115.5 x 4.27 = 493.5 kgm/s

(d) required relation is

Momentum of center of mass = sum of the mometum of two sprinters

P = P1 + P2

MV = m1v1 + m2v2

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