Momentum and Speed?

Question

Two ice skaters, with masses of 50 kg and 75 kg, are at the center of a 60 m diameter circular rink. The skaters push off against each other and glide to opposite edges of the rink. If the heavier skater reaches the edge in 14 s, how long does the lighter skater take to reach the edge?

I have no idea why a cannot get this one. Here's what I am doing:

d = Vit + 1/2at^2

30=1/2(a)(14)^2
a = 0.306 m/s^2 <--acceleration of heavier skater


ma = ma
75(0.306) = 50a
a = 0.459 m/s <--acceleration of lighter skater

30 = 1/2(0.459)t^2
t = 11.430 sec

^That answer appears to be incorrect. Any suggestions?


Thanks

Answer 1

Remember the momentum is conserved.

In the frame of reference of the rink, both the skaters are at rest initially. The initial total momentum is therefore zero.

After the push heavier skater gets a speed of 60/14 meters per second in a certain direction. His velocity will be in the direction of motion and its magnitude would be 60/14 m/s.

By Newtons third law of motion the action on the other skater will have to be equal and opposite. So let us say that his velocity is in the opposite direction of that of the heavy skater and the magnitude say 'v'

Then the total momentum after the push = 75 * 60/14 - 50 * v

( minus sign because the lighter skater Travers in the opposite direction to the heavy one)

As momentum is conserved it should be equal to the initial total momentum which is zero.

Hence 75 * 60/14 - 50 * v =0

therefore v = (75*60/14)/50 = (3*60)/(2 *14)

time taken =60/v = (60 * 2 *14) /(3*60) = 28/3 = 9.33 s

Answer 2

You can check by using the momentum equation:
Let m1=75 and m2=50.
m1v1 = m2v2. Since v1 = r/t1 and v2=r/v2, you have
m1r/t1 = m2r/t2, or t1/t2 = m2/m1.
So t1 = 14 * 50/75 = 9.333 sec.
The answer you got using d = 1/2 at^2 assumes acceleration is continuous, but it's really only present during the pushoff (a short time), which I think you can assume is permitted to be approximated as instantaneous, since no information is given about the duration or distance over which the acceleration acts.

Answer 3

First solve for speed of heavier skater:

speed=distance/time
=(60/2)/14
=2.14m/s


By ratio and proportion solve speed of lighter skater, v.

75:2.14=50:v
v=50*2.14/75
=1.43m/s

Time of lighter skater=30/1.43
=21s

Answer 4

the initial velocity is not zero..