Two skaters collide an embrace, in a completely inelastic collision. That is, they stick together after impact. The origin is placed at the point of collision. Alfred, whose mass is 83 kg, is originally moving east with speed 6.2 km/h. Barbara, whose mass is 55 kg, is originally moving north with speed 7.8 km/h. What is the velocity v of the couple after impact? Hint: You must find vx, vy and the angle theta.

Question

Two skaters collide an embrace, in a completely inelastic collision. That is, they stick together after impact. The origin is placed at the point of collision. Alfred, whose mass is 83 kg, is originally moving east with speed 6.2 km/h. Barbara, whose mass is 55 kg, is originally moving north with speed 7.8 km/h. What is the velocity v of the couple after impact? Hint: You must find vx, vy and the angle theta.

Answer 1

Darlene/Marie/Anna/Meagan -- please use the same name for your posts.

Answer 2

momentum is conserved in both the x and y directions

x ... east ... 83 * 6.2 = (83 + 55) vx

y ... north ... 55 * 7.8 = (55 + 83) vy

v^2 = vx^2 + vy^2

tan(Θ) = vy / vx

so...you're schizophrenic?
(multiple personalities)

Answer 3

To solve this problem, we can break down the velocities of Alfred and Barbara into their x and y components.

Given:
Mass of Alfred (m1) = 83 kg
Velocity of Alfred before collision (v1) = 6.2 km/h
Mass of Barbara (m2) = 55 kg
Velocity of Barbara before collision (v2) = 7.8 km/h

Step 1: Convert the velocities from km/h to m/s.
v1 = 6.2 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 1.72 m/s
v2 = 7.8 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 2.17 m/s

Step 2: Determine the x and y components of the velocities.
The x-component of v1 is v1x = v1 * cos(theta1), where theta1 is the angle between the velocity vector and the x-axis.
The y-component of v2 is v1y = v1 * sin(theta1), where theta1 is the angle between the velocity vector and the y-axis.
Similarly, define theta2 for Barbara.

Step 3: Find the x and y components of the velocity of the couple after the collision.
Using the principle of conservation of momentum:
(m1 * v1x + m2 * v2x) = (m1 + m2) * vx
(m1 * v1y + m2 * v2y) = (m1 + m2) * vy

Step 4: Substitute the given values and solve for vx and vy.
(m1 * v1x + m2 * v2x) = (m1 + m2) * vx
(83 kg * v1 * cos(theta1) + 55 kg * 0) = (83 kg + 55 kg) * vx
(83 kg * 1.72 m/s * cos(theta1)) = (138 kg) * vx

(m1 * v1y + m2 * v2y) = (m1 + m2) * vy
(83 kg * 0 + 55 kg * v2 * sin(theta2)) = (83 kg + 55 kg) * vy
(55 kg * 2.17 m/s * sin(theta2)) = (138 kg) * vy

Step 5: Divide the two equations to eliminate vx and vy.
(83 kg * 1.72 m/s * cos(theta1)) / (55 kg * 2.17 m/s * sin(theta2)) = (83 kg + 55 kg) * vx / (138 kg) * vy

Simplify the equation:
1.29 = vx / vy

Step 6: Use the Pythagorean theorem to find the magnitude of the velocity v after impact.
v = sqrt(vx^2 + vy^2)

Step 7: Find the angle theta.
theta = arctan(vy / vx)

Step 8: Substitute the value of vx / vy from Step 5 into Step 7.
theta = arctan(1.29)

Now, we have the magnitude of the velocity v and the angle theta of the couple after impact.

Answer 4

To find the velocity v of the couple after impact, we first need to break down the initial velocities of Alfred and Barbara into their respective x and y components.

Given:
- Alfred's mass (mA) = 83 kg
- Alfred's initial velocity (vA) = 6.2 km/h
- Barbara's mass (mB) = 55 kg
- Barbara's initial velocity (vB) = 7.8 km/h

Step 1: Convert the initial velocities from km/h to m/s.
- Alfred's initial velocity (vA) = 6.2 km/h * (1000 m/1 km) * (1 h/3600 s) = 1.72 m/s
- Barbara's initial velocity (vB) = 7.8 km/h * (1000 m/1 km) * (1 h/3600 s) = 2.17 m/s

Step 2: Find the x and y components of the initial velocities.
- The x component (vAx) of Alfred's initial velocity (vA) is the velocity in the east direction and is equal to vA since he is moving directly east.
- vAx = vA = 1.72 m/s

- The y component (vAy) of Alfred's initial velocity (vA) is the velocity in the north direction and is equal to 0 since he is not moving in the north direction.
- vAy = 0 m/s

- The x component (vBx) of Barbara's initial velocity (vB) is 0 since she is not moving in the east direction.
- vBx = 0 m/s

- The y component (vBy) of Barbara's initial velocity (vB) is the velocity in the north direction and is equal to vB.
- vBy = vB = 2.17 m/s

Step 3: Use the principles of conservation of momentum to find the x and y components of the final velocity (vx and vy).

Conservation of momentum in the x-direction:
mAx * vAx + mBx * vBx = (mAx + mBx) * vx
Since mBx = 0,
mAx * vAx = (mAx + 0) * vx
vAx = vBx = 0, so this equation is not helpful in finding vx.

Conservation of momentum in the y-direction:
mAy * vAy + mBy * vBy = (mAy + mBy) * vy
Since vAy = 0,
0 + mBy * vBy = (0 + mBy) * vy
mBy * vBy = mBy * vy
vBy = vy

Therefore, vy = 2.17 m/s

Step 4: Use trigonometry to find the magnitude (v) and angle (θ) of the final velocity.
v = √(vx^2 + vy^2)

Given that vy = 2.17 m/s,
v = √(vx^2 + (2.17)^2)
v^2 = vx^2 + (2.17)^2

Now, we need to find the angle θ.
tan(θ) = vy / vx
tan(θ) = 2.17 / vx

Step 5: Substitute the expression for vx from the equation in Step 4 into the equation in Step 5 to solve for θ.

tan(θ) = 2.17 / √(vx^2 + (2.17)^2)

Solve this equation for θ and find the value.

By plugging in the values obtained for vy and vx in the equations above, we can solve for v and θ to find the final velocity (v) and its angle (θ) of the couple after impact.