A 49.5-kg skater is traveling due east at a speed of 2.88 m/s. A 72.5-kg skater is moving due south at a speed of 6.82 m/s. They collide and hold on to each other after the collision, managing to move off at an angle south of east, with a speed of vf.

Question

A 49.5-kg skater is traveling due east at a speed of 2.88 m/s. A 72.5-kg skater is moving due south at a speed of 6.82 m/s. They collide and hold on to each other after the collision, managing to move off at an angle south of east, with a speed of vf.

(a) Find the angle.

(b) Find the speed vf, assuming that friction can be ignored.

I was able to get part A which is 73.9167 ° , but I am not sure how to do part B or even where to start.

Answer 1

well, you must have found the east and south speeds to get the angle from

tan angle = speed south/speed east

then
Vf = sqrt (Vsouth^2 + Veast^2)

Answer 2

A 20kg box is pulled by a horizontal wire on a circle on rough horizontal for which the coefficient on the kinetic is 0.350.calculate the work down by friction during one complete circle trip if the radiudis 3m.

Answer 3

To solve part B of the problem, we need to use the principles of conservation of momentum and conservation of kinetic energy.

First, let's consider the initial momentum before the collision. Each skater has both magnitude and direction associated with their momentum. The magnitude of the momentum is given by the product of mass and velocity, while the direction is indicated by the angle of travel.

For the 49.5-kg skater traveling due east at 2.88 m/s, the initial momentum is calculated as follows:

Momentum1 = (mass1)(velocity1) = (49.5 kg)(2.88 m/s) = 142.56 kg⋅m/s, directed east (0°).

For the 72.5-kg skater traveling due south at 6.82 m/s, the initial momentum is calculated as follows:

Momentum2 = (mass2)(velocity2) = (72.5 kg)(6.82 m/s) = 494.15 kg⋅m/s, directed south (270°).

Now, let's analyze the final momentum after the collision. Since the skaters hold on to each other after the collision, they move off at an angle south of east with a combined speed of vf.

We can break down the final momentum into its x-axis and y-axis components. Let's define the x-axis to be east and the y-axis to be north.

The x-component of the final momentum (Momentumf_x) is given by:

Momentumf_x = (mass1 + mass2) × vf × cos(135°).

The y-component of the final momentum (Momentumf_y) is given by:

Momentumf_y = (mass1 + mass2) × vf × sin(135°).

Using the principles of conservation of momentum, we know that the x-component of the final momentum (Momentumf_x) must equal the initial momentum in the x-direction (Momentum1). Similarly, the y-component of the final momentum (Momentumf_y) must equal the initial momentum in the y-direction (Momentum2).

Now, let's solve for vf:

Momentumf_x = Momentum1
(mass1 + mass2) × vf × cos(135°) = Momentum1
vf × cos(135°) = Momentum1 / (mass1 + mass2)

Momentumf_y = Momentum2
(mass1 + mass2) × vf × sin(135°) = Momentum2
vf × sin(135°) = Momentum2 / (mass1 + mass2)

Dividing these two equations gives us the tangent of the angle:

tan(angle) = (Momentum2 / (mass1 + mass2)) / (Momentum1 / (mass1 + mass2))
= Momentum2 / Momentum1

Now we can calculate the angle:

angle = arctan(Momentum2 / Momentum1)

Finally, to find vf, we can substitute the value of angle into either of the previous equations:

vf = (Momentum1 / (mass1 + mass2)) / cos(angle)

Plug in the given values and calculate vf.