A speed skater moving across frictionless ice at 8.70 hits a 4.20-m-wide patch of rough ice. She slows steadily, then continues on at 5.90
To solve this problem, we can use the principle of conservation of momentum.
1. Calculate the initial momentum of the skater before hitting the rough ice patch.
Momentum (p) = mass (m) × velocity (v)
Initial momentum (p1) = m × v
Since the mass is not given, we can assume it cancels out in this case, so we can focus on velocities.
Initial velocity (v1) = 8.70 m/s
2. Calculate the final momentum of the skater after hitting the rough ice patch.
Final momentum (p2) = m × v
Final velocity (v2) = 5.90 m/s
3. Apply the principle of conservation of momentum.
The initial momentum (p1) will equal the final momentum (p2).
p1 = p2
m × v1 = m × v2
v1 = (m × v2) / m
Since the mass cancels out, the initial and final velocities are equal.
So, the skater will continue to move at a velocity of 5.90 m/s after hitting the rough ice patch.
To find the acceleration of the speed skater as she slows down on the rough ice patch, we need to use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity of the skater (5.90 m/s)
u = initial velocity of the skater (8.70 m/s)
a = acceleration of the skater
s = distance covered by the skater on the rough ice patch (4.20 m)
Rearranging the equation to solve for acceleration (a), we get:
a = (v^2 - u^2) / (2s)
Substituting the given values into the equation, we have:
a = ( (5.90 m/s)^2 - (8.70 m/s)^2 ) / (2 * 4.20 m)
Now, let's calculate the value of acceleration using this equation.
a = (34.81 m^2/s^2 - 75.69 m^2/s^2) / (8.40 m)
a = -40.88 m^2/s^2 / 8.40 m
Calculating the acceleration gives us:
a = -4.87 m/s^2
Therefore, the acceleration of the speed skater as she slows down on the rough ice patch is approximately -4.87 m/s^2 (negative sign indicates deceleration).