Two ice skaters, with masses of 50 kg and 75 kg, are at the center of a 60-m-dia

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 16 s, how long does the lighter skater take to reach the edge?

A 14 g bullet is fired into a 14 kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 7.6 cm across the table. What was the speed of the bullet? (Assume that the coefficient of kinetic friction between the block and the table is 0.20.)

Explanation / Answer

(a) In the frame of reference of the rink, both the skaters are at rest initially. Therefore, their initial total momentum is zero.

After the push heavier skater gets a speed of 60/16 meters per second in a certain direction.

His velocity will be in the direction of motion and its magnitude would be 60/16 m/s = 3.75 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 * 3.75) - 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 * 3.75) - 50 * v =0

Therefore v = (75*3.75)/50 = 5.625 m/s

Time taken = 60/v = 60/5.625 = 10.667 seconds

(b) Given,

Mass of the bullet is m = 14g = 0.014 kg

Mass of the block is M = 14 kg

Mass of block plus the bullet = (M+m) = 14+0.014 = 14.014 kg

Let V1 be the velocity of the bullet and V2 be the combined velocity of bullet plus the block after impact

By law of conservation of momentum,

0.014 * V1 = 14.014 * V2 ---- (1)

Now,

Frictional force on the bullet block system = coefficient of friction * mass * acceleration due to gravity

= 0.20 * 14.014 * 9.80665 N

==> Deceleration = a = 0.20 * 9.80665 m/s^2

Therefore,

V22 = (2)*(0.076 m)*(0.2)*(9.80665 m/s^2) = 0.298224 (m/s)^2

Now, using (1), we get

==> V1 = (14.014/0.014)*[(2)(0.076 m)(0.2)(9.80665 m/s^2)]1/2

==> Speed of the bullet is = V1 = 546.5516 m/s

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