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?
To solve this problem, we can use the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is calculated by multiplying its mass by its velocity. So, we can calculate the momentum of each skater before and after they push off against each other.
Let's denote the mass of the first skater as m1 and the mass of the second skater as m2.
Before the push-off, both skaters are at rest, so their initial momentum is zero.
After the push-off, the first skater with a mass of 61 kg acquires a speed of 0.63 m/s. Therefore, his momentum is given by:
Momentum1 = m1 * velocity1 = (61 kg) * (0.63 m/s)
Similarly, the second skater has a velocity of 0.86 m/s, so his momentum is:
Momentum2 = m2 * velocity2 = (m2) * (0.86 m/s)
According to the law of conservation of momentum, the total momentum before and after the collision should be the same. Therefore:
Momentum1 + Momentum2 = 0
((61 kg) * (0.63 m/s)) + (m2 * (0.86 m/s)) = 0
Now, we can solve this equation to find the mass of the second skater m2.
(38.43 kg*m/s) + (0.86 m/s * m2) = 0
0.86 m/s * m2 = -38.43 kg*m/s
m2 = -38.43 kg*m/s / 0.86 m/s
m2 = -44.77 kg
Since mass cannot be negative, we discard the negative value.
Hence, the mass of the second skater is approximately 44.77 kg.