Let's tackle these physics problems one by one, breaking them down into manageable parts for clarity and understanding. Each question involves concepts of relative velocity, direction, and circular motion, which are fundamental in physics.

Question 1: Rugby Player's Pass

In this scenario, we need to determine the smallest angle at which the rugby player can pass the ball while ensuring the ball's velocity relative to the field does not have a positive x-component. The player runs at a speed of 3.7 m/s along the positive x-axis, and he passes the ball with a speed of 5.2 m/s relative to himself.

Step-by-Step Analysis

To find the angle, we can use vector components. The velocity of the ball relative to the field can be expressed as:

• V_ball_x = V_player + V_BP * cos(θ)
• V_ball_y = V_BP * sin(θ)

Where:

• V_player = 3.7 m/s (the player's speed)
• V_BP = 5.2 m/s (the ball's speed relative to the player)
• θ is the angle of the pass.

For the pass to be legal, the x-component of the ball's velocity must be less than or equal to zero:

3.7 m/s + 5.2 m/s * cos(θ) ≤ 0

Rearranging gives:

cos(θ) ≤ -3.7 / 5.2

Calculating the right side:

cos(θ) ≤ -0.7115

Now, we find the angle θ:

θ = cos⁻¹(-0.7115)

Calculating this gives:

θ ≈ 134.0°

This means the smallest angle at which the player can legally pass the ball is approximately 134 degrees from the positive x-axis.

Question 2: Ships A and B

Now, let's analyze the motion of the two ships. Ship A travels northwest at 25 knots, while Ship B travels at 35 knots, 40° west of south. We need to find the relative velocity, the distance apart after a certain time, and the bearing of Ship B relative to Ship A.

(a) Magnitude of Velocity of Ship A Relative to B

First, we convert the directions into vector components:

• Ship A:
  • V_Ax = -25 * cos(45°) = -17.68 knots
  • V_Ay = 25 * sin(45°) = 17.68 knots
• Ship B:
  • V_Bx = -35 * sin(40°) = -22.49 knots
  • V_By = -35 * cos(40°) = -26.83 knots

Now, the relative velocity of A with respect to B is:

V_AB = V_A - V_B

Calculating the components:

• V_ABx = -17.68 - (-22.49) = 4.81 knots
• V_ABy = 17.68 - (-26.83) = 44.51 knots

The magnitude of the relative velocity is:

|V_AB| = √(4.81² + 44.51²) ≈ 44.56 knots

(b) Direction of Velocity of Ship A Relative to B

To find the direction, we calculate the angle using:

tan(φ) = V_ABy / V_ABx

Calculating gives:

φ = tan⁻¹(44.51 / 4.81) ≈ 83.5°

(c) Time Until Ships are 135 Nautical Miles Apart

Using the relative speed:

Distance = Speed × Time

Setting up the equation:

135 = 44.56 × t

Solving for t gives:

t ≈ 3.03 hours

(d) Bearing of B Relative to A

After 3.03 hours, we can find the position of Ship B relative to Ship A. The angle can be calculated using the relative velocity components:

tan(θ) = V_ABy / V_ABx

Thus, the bearing of B relative to A is:

θ ≈ 83.5° west of north

Question 3: Airplane Speed Relative to Air

In this problem, we need to find the speed of the airplane relative to the air after flying for 15 minutes in a wind blowing at 42 km/h at an angle of 34° south of east. The airplane is over a town that is 40 km due north of the starting point.

Breaking Down the Components

First, we convert the wind's velocity into components:

• V_wind_x = 42 * cos(34°) ≈ 34.5 km/h
• V_wind_y = -42 * sin(34°) ≈ -14.2 km/h

Next, we need to find the airplane's velocity components. The airplane's displacement is 40 km north, and it took 15 minutes (0.25 hours) to reach that point:

V_airplane_y = 40 km / 0.25 h = 160 km/h

Since the airplane is moving north, its x