Two skaters of mass m1=50 kg and m2=70 kg are standing motionless on a horizontal ice surface. They are initially a distance L= 8.0 meters apart. They hold a massless rope between them. After pulling the rope, the skater of mass m1 has moved a distance l= 2.0 meters away from his initial position. We can completely neglect friction in this problem.
What is the distance L′ between the two skaters when the skater of mass m1 has moved a distance l? (in meters)
To find the distance L' between the two skaters when the skater of mass m1 has moved a distance l, we can use the principles of conservation of momentum and conservation of mass.
Given:
Mass of skater 1 (m1) = 50 kg
Mass of skater 2 (m2) = 70 kg
Initial distance between skaters (L) = 8.0 meters
Distance skater 1 moved (l) = 2.0 meters
Since the skaters are initially motionless, the total initial momentum is zero. After the skater 1 moves a distance l, the total momentum remains zero.
To solve for L', we need to consider the mass ratio between skater 1 and skater 2:
m1/m2 = L'/L
Substituting the given values:
50 kg / 70 kg = L' / 8.0 meters
Now we can solve for L':
L' = (50 kg * 8.0 meters) / 70 kg
L' = 400 kg*m / 70 kg
L' ≈ 5.714 meters
Therefore, the distance L' between the two skaters when skater 1 has moved a distance l is approximately 5.714 meters.
To find the distance L' between the two skaters when the skater of mass m1 has moved a distance l, we can use the concept of conservation of momentum.
Let's define v1 as the final velocity of skater m1 and v2 as the final velocity of skater m2. Since there is no external force acting on the system, the total momentum before and after the pull should be the same.
Initially, the total momentum is zero since both skaters are motionless. Therefore, the total momentum after the pull should also be zero.
The momentum of skater m1 after the pull is given by p1 = m1 * v1, where m1 is the mass of skater m1.
The momentum of skater m2 after the pull is given by p2 = m2 * v2, where m2 is the mass of skater m2.
Since the total momentum is zero, we can write the equation:
p1 + p2 = 0
m1 * v1 + m2 * v2 = 0
Since the skaters move in opposite directions, the velocities have opposite signs. Therefore, we can rewrite the equation as:
m1 * v1 = -m2 * v2
Now, let's use the equation of conservation of total mechanical energy. The distance L' between the skaters is related to their initial positions and the distance the skater m1 has moved:
L' = L - l,
where L is the initial distance between the skaters, and l is the distance the skater m1 has moved.
Therefore, we can rewrite the equation as:
m1 * v1 = -m2 * v2 * (L - l) / L
Now, we can substitute the given values into the equation:
m1 = 50 kg
m2 = 70 kg
L = 8.0 meters
l = 2.0 meters
Substituting these values, we get:
50 kg * v1 = -70 kg * v2 * (8.0 m - 2.0 m) / 8.0 m
50 kg * v1 = -70 kg * v2 * 6.0 m / 8.0 m
Simplifying the equation:
50 kg * v1 = -70 kg * v2 * 0.75
Simplifying further:
v1 = -1.05 * v2
Now, we need to find the distance L' between the skaters when the skater of mass m1 has moved a distance l. We can use the equation L' = L - l:
L' = 8.0 meters - 2.0 meters
L' = 6.0 meters
Therefore, the distance L' between the two skaters when the skater of mass m1 has moved a distance l is 6.0 meters.
The cneter of gravity has to remain at the same point.
measuring from the center of gravity point, then if d is the distance the m1 is frm the cg, then
50d=70(8-d) and after the jerk, then we know
50(d-2)=70(L'-(d-2)
you have two equations, two unknowns
solve the first equation for d
50d=560-70d
120d=560
d= 4 2/3
now the second equation..
50(d-2)=70(L'-d+2)
50(2 2/3)=70L' -70(4 2/3-2)
L'= 120(2 2/3)/70=4.7 m check my math, done in my head.