A 45.0-kg skater is traveling due east at a speed of 2.85 m/s. A 72.5-kg skater is moving due south at a speed of 7.35 m/s. They collide and hold on to each other after the collision, managing to move off at an angle θ south of east, with a speed of vf. Find the following.
(a) the angle θ
°
(b) the speed vf, assuming that friction can be ignored
m/s
south momentum = 75 * 7.35 = 551
east momentum = 45 * 2.85 = 128
new mass = 45 + 75 = 120
new momentum = old momentum (Newton #1 law)
south speed = 551/120 = 4.59
east speed = 128/120 = 1.07
theta = tan^-1 (4.59/1.07) = 76.9 degrees
Vf = sqrt(4.59^2+1.07^2) = 4.71 m/s
Answer
(a) Well, it sounds like these skaters are having a bit of an "oops" moment. It's like two clumsy penguins colliding on an ice rink! Anyway, let's calculate the angle θ.
To find θ, we can use some good old trigonometry. We need to find the tangent of θ. Since we have the opposite and adjacent sides of the triangle formed by the skaters' velocities, we can use the formula tangent θ = opposite/adjacent.
In this case, the opposite side is the southward velocity of the second skater, which is 7.35 m/s. The adjacent side is the eastward velocity of the first skater, which is 2.85 m/s.
So, tangent θ = 7.35 m/s / 2.85 m/s.
Using a calculator, we find that θ ≈ 69.213°.
So, the angle θ is approximately 69.213°.
(b) Now, let's find the final speed, vf.
Before the collision, we can use the law of conservation of momentum, which tells us that the total momentum before the collision is equal to the total momentum after the collision.
The initial momentum of the 45.0-kg skater is given by: momentum1 = mass1 * velocity1 = 45.0 kg * 2.85 m/s.
The initial momentum of the 72.5-kg skater is given by: momentum2 = mass2 * velocity2 = 72.5 kg * 7.35 m/s.
The total momentum before the collision is the sum of these two:
total momentum before = momentum1 + momentum2.
After the collision, the skaters move off at an angle θ south of east with a speed vf. So, we can calculate the final momentum:
total momentum after = (mass1 + mass2) * vf.
Since the total momentum before and after the collision must be the same, we have:
momentum1 + momentum2 = (mass1 + mass2) * vf.
Plugging in the given values, we can solve for vf:
(45.0 kg * 2.85 m/s) + (72.5 kg * 7.35 m/s) = (45.0 kg + 72.5 kg) * vf.
Using a calculator, we find that vf ≈ 5.75 m/s.
So, the speed vf, assuming that friction can be ignored, is approximately 5.75 m/s.
Whoa, that's quite a "wheee" moment for our skaters! I hope they're okay and remembered to let out a good ol' "ta-da!" at the end of their unexpected performance.
To find the angle θ and the speed vf after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.
(a) To find the angle θ, we can use the concept of vector addition. Let's assume that the initial direction of the 45.0-kg skater is the x-axis, and the initial direction of the 72.5-kg skater is the y-axis.
The x-component of the momentum before the collision is given by:
P1x = mass1 * velocity1x = 45.0 kg * 2.85 m/s = 128.25 kg·m/s
The y-component of the momentum before the collision is given by:
P2y = mass2 * velocity2y = 72.5 kg * (-7.35 m/s) = -533.175 kg·m/s
The total x-component of momentum after the collision is given by:
Pfx = (mass1 + mass2) * vfx = (45.0 kg + 72.5 kg) * vf * cos(θ)
The total y-component of momentum after the collision is given by:
Pfy = (mass1 + mass2) * vfy = (45.0 kg + 72.5 kg) * vf * sin(θ)
Using the principle of conservation of momentum, we can equate the initial and final momenta in both the x and y directions:
P1x = Pfx
P2y = Pfy
Substituting the values:
128.25 kg·m/s = (45.0 kg + 72.5 kg) * vf * cos(θ)
-533.175 kg·m/s = (45.0 kg + 72.5 kg) * vf * sin(θ)
Dividing the second equation by the first equation:
-533.175 kg·m/s / 128.25 kg·m/s = sin(θ) / cos(θ)
Solving for θ:
tan(θ) = -533.175 kg·m/s / 128.25 kg·m/s
θ = atan(-533.175 kg·m/s / 128.25 kg·m/s)
θ ≈ -77.2°
Therefore, the angle θ is approximately -77.2°.
(b) To find the speed vf, we can use the principle of conservation of kinetic energy. The initial kinetic energy is given by:
KE_initial = 0.5 * mass1 * velocity1^2 + 0.5 * mass2 * velocity2^2
KE_initial = 0.5 * 45.0 kg * (2.85 m/s)^2 + 0.5 * 72.5 kg * (7.35 m/s)^2
KE_initial ≈ 1004.9 J
The final kinetic energy is given by:
KE_final = 0.5 * (mass1 + mass2) * vf^2
Using the principle of conservation of kinetic energy:
KE_initial = KE_final
1004.9 J = 0.5 * (45.0 kg + 72.5 kg) * vf^2
1004.9 J = 0.5 * 117.5 kg * vf^2
Solving for vf:
vf^2 = (1004.9 J) / (0.5 * 117.5 kg)
vf^2 ≈ 17.0851 m^2/s^2
vf = sqrt(17.0851 m^2/s^2)
vf ≈ 4.134 m/s
Therefore, the speed vf, assuming friction can be ignored, is approximately 4.134 m/s.
To find the angle θ and the speed vf, we can apply the principles of conservation of momentum and conservation of kinetic energy. Here's how you can solve the problem step by step:
Step 1: Determine the initial momentum of each skater in the x and y directions.
The initial momentum in the x direction is given by:
P₁x = m₁ * v₁x (for skater 1)
P₂x = m₂ * v₂x (for skater 2)
Given:
m₁ = 45.0 kg (mass of skater 1)
v₁x = 2.85 m/s (eastward velocity of skater 1)
m₂ = 72.5 kg (mass of skater 2)
v₂x = 0 m/s (since skater 2 is moving due south)
P₁x = (45.0 kg) * (2.85 m/s) = 128.25 kg·m/s
P₂x = (72.5 kg) * (0 m/s) = 0 kg·m/s
The initial momentum in the y direction is given by:
P₁y = m₁ * v₁y (for skater 1)
P₂y = m₂ * v₂y (for skater 2)
Given:
v₁y = 0 m/s (since skater 1 is moving due east)
v₂y = -7.35 m/s (southward velocity of skater 2)
P₁y = (45.0 kg) * (0 m/s) = 0 kg·m/s
P₂y = (72.5 kg) * (-7.35 m/s) = -533.625 kg·m/s
Step 2: Calculate the total initial momentum in the x and y directions.
P_initial_x = P₁x + P₂x = 128.25 kg·m/s + 0 kg·m/s = 128.25 kg·m/s
P_initial_y = P₁y + P₂y = 0 kg·m/s + (-533.625 kg·m/s) = -533.625 kg·m/s
Step 3: Determine the final momentum in the x and y directions.
Since the skaters hold on to each other, the final momentum in the x and y directions are the same.
P_final_x = P_final_y = P_final
Step 4: Apply conservation of momentum to find P_final.
Conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.
P_initial = P_final
P_initial = √((P_initial_x)² + (P_initial_y)²)
P_final = √((P_final_x)² + (P_final_y)²)
√((P_initial_x)² + (P_initial_y)²) = √((P_final_x)² + (P_final_y)²)
(P_initial_x)² + (P_initial_y)² = (P_final_x)² + (P_final_y)²
(128.25 kg·m/s)² + (-533.625 kg·m/s)² = (P_final_x)² + (P_final_y)²
16484.56 kg²·m²/s² + 284707.8906 kg²·m²/s² = (P_final_x)² + (P_final_y)²
301192.4506 kg²·m²/s² = (P_final_x)² + (P_final_y)²
Step 5: Solve for P_final.
P_final = √(301192.4506 kg²·m²/s²) = 548.9 kg·m/s
Step 6: Determine the angle θ.
tan θ = P_final_y / P_final_x
θ = tan⁻¹(P_final_y / P_final_x)
θ = tan⁻¹(-533.625 kg·m/s / 128.25 kg·m/s) = -74.76°
Note: The negative sign indicates the angle is measured below the positive x-axis, which is south of east.
(a) The angle θ ≈ -74.76° (south of east)
(b) The speed vf = P_final = 548.9 kg·m/s