Two ice skaters stand at rest in the center of an ice rink. When they push off against one another the 61-kg skater acquires a speed of 0.63 m/s .

If the speed of the other skater is 0.86 m/s , what is this skater's mass?

32. Two ice skaters stand at rest in the center of an ice rink. When they push off against each 45-kg skater acquires a speed of 0.62 m/s. If the speed of the other skater is 0.89 m/s, what is that skater's mass?

Well, well, well, look at these ice-skating buddies trying to calculate their mass! It's like they're playing a fun little physics game on the ice. Let's see if I can help them out.

Now, according to the laws of conservation of momentum, the total momentum before and after the push-off should be the same. We have one skater with a mass of 61 kg and a velocity of 0.63 m/s, and another skater with an unknown mass but a velocity of 0.86 m/s.

To calculate the mass of the second skater, we can use this equation:

(mass of skater 1 × velocity of skater 1) + (mass of skater 2 × velocity of skater 2) = 0

We know the values for the first skater, so we can substitute them into the equation:

(61 kg × 0.63 m/s) + (mass of skater 2 × 0.86 m/s) = 0

Now, to solve for the mass of the second skater:

61 kg × 0.63 m/s = (mass of skater 2 × 0.86 m/s)

Divide both sides by 0.86 m/s:

(61 kg × 0.63 m/s) / 0.86 m/s = mass of skater 2

And the answer pops out:

Mass of skater 2 = 44.674 kg (approximately)

So, dear skater friend, it looks like your mass is approximately 44.674 kg! Now go out there and continue having a blast on the ice!

To solve this problem, we can use the principle of conservation of momentum. In an isolated system, the total momentum before an event is equal to the total momentum after the event.

Given:
Mass of the first skater (m1) = 61 kg
Speed of the first skater (v1) = 0.63 m/s
Speed of the second skater (v2) = 0.86 m/s

Let the mass of the second skater be denoted as m2.

Using the conservation of momentum, we can write the equation:

m1 * v1 + m2 * v2 = 0

Substituting the given values into the equation, we get:

61 * 0.63 + m2 * 0.86 = 0

Simplifying the equation:

38.43 + 0.86m2 = 0

0.86m2 = -38.43

Dividing both sides by 0.86:

m2 = -38.43 / 0.86

m2 ≈ -44.75 kg

Since mass cannot be negative, this indicates an error in the calculation or setup of the problem. Please check the values given and re-verify the problem setup.

To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the skaters push off against each other is equal to the total momentum after the push-off.

Let's assume that the mass of the second skater is 'm' kg.

Before the push-off, the first skater's momentum is zero since the skater is at rest. The second skater's momentum can be calculated using the formula: momentum = mass × velocity. Therefore, the second skater's momentum is (m × 0.86) kg.m/s.

After the push-off, the total momentum is the sum of the momenta of the two skaters. The first skater's momentum is (61 × 0.63) kg.m/s, and the second skater's momentum is (m × 0.86) kg.m/s.

According to the principle of conservation of momentum, the total momentum before and after the push-off should be the same. Therefore, we can set up the following equation:

0 = (61 × 0.63) + (m × 0.86)

Now we can solve this equation to find the value of 'm':

0 = 38.43 + 0.86m

Rearranging the equation, we have:

0.86m = -38.43

Dividing both sides by 0.86:

m = -38.43 / 0.86

m ≈ -44.7

Since mass cannot be negative, we can conclude that the mass of the second skater is approximately 44.7 kg.

momentum is zero before so zero after.

61*.63 + m* -.86 = 0
note speeds are opposite so velocity of one or the other is negative

m = 61 (.63/.86)