At the center of a 41.0 m diameter circular ice rink, a 61.0 kg skater traveling north at 1.05 m/s collides with and holds onto a 77.0 kg skater who had been heading west at 1.90 m/s. How long (in seconds) will it take them to reach the edge of the rink assuming negligible friction and drag?
use conservation of momentum to find the new momentum.
New momentum: 61*1.05N+77*1.90W
absolute speed: sqrt [(61*1.05)^2 + (77*1.9)^2 ]
then time= distance/speed= 41/2 * 1/speed
I did this and it still says it is incorrect...
I bet you made an error. Do you see anything wrong with the approach, or thinking? Have you checked your work? Post it here, and someone can critique it.
I used the equation m1vf1 + m2vf2 = m1vi1 + m2vi2 and got a vf of 1.52m/s
I then plugged that into the equation t=d/v => t=20.5/1.52 and got 13.5s
Is there some reason you did not do what I told you? What you did is really very wrong. Please look at the solution I outlined. And I see I have an error, where I typed absolute speed, it should read absolute momentum, then a new line
absolute speed= absolute mometnum/total mass I get about 1.1 for the speed (not 1.1).
Yeah I did do it the way you told me at first and got a number that made absolutely no sense but the way you just illustrated makes more sense....thank you
To find the time it takes for the skaters to reach the edge of the rink, we need to consider the conservation of momentum in this system.
Let's break down the problem into two components: the momentum in the x-direction and the momentum in the y-direction.
First, let's find the initial momentum in the x-direction before the collision:
Given:
Mass of skater 1 (m1) = 61.0 kg
Velocity of skater 1 in the x-direction (v1x) = 1.90 m/s (west)
Momentum of skater 1 in the x-direction before the collision (p1x_initial) = m1 * v1x
Next, let's find the initial momentum in the y-direction before the collision:
Given:
Mass of skater 2 (m2) = 77.0 kg
Velocity of skater 2 in the y-direction (v2y) = 1.05 m/s (north)
Momentum of skater 2 in the y-direction before the collision (p2y_initial) = m2 * v2y
Since we're assuming no external forces, momentum is conserved. Therefore, the total initial momentum in the x-direction (p_initial_x) and the total initial momentum in the y-direction (p_initial_y) can be found as follow:
p_initial_x = p1x_initial
p_initial_y = p2y_initial
Now, let's find the final momentum in the x-direction after the collision:
The skaters collide and hold onto each other, resulting in a shared velocity in the x-direction.
Given:
Mass of the combined skaters (m_total) = m1 + m2
Since momentum is conserved, the final momentum in the x-direction (p_final_x) can be calculated as:
p_final_x = p_initial_x
Now let's find the final momentum in the y-direction after the collision:
The combined skaters continue to move in the y-direction with their initial momentum.
Since momentum is conserved, the final momentum in the y-direction (p_final_y) can be calculated as:
p_final_y = p_initial_y
Now that we have the final momentum in both the x-direction and y-direction, we can calculate the combined velocity of the skaters using the Pythagorean theorem:
Given:
p_final_x = p_initial_x
p_final_y = p_initial_y
The combined velocity of the skaters (v_combined) can be calculated as:
v_combined = sqrt(p_final_x^2 + p_final_y^2) / m_total
Finally, we can determine the time it takes for the skaters to reach the edge of the rink. The skaters start at the center of the circular rink and must travel a distance equal to the radius (which is half the diameter).
Given:
Radius of the rink (r) = 41.0 m
The time it takes for the skaters to reach the edge of the rink (t) can be calculated using the formula:
t = r / v_combined
Substituting the values we've calculated, we can find the time it takes for the skaters to reach the edge of the rink.