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Write the momentum vector of a 10 kg ball moving with velocity 40 m/s SE (-45 deg) in standard unit vector notation. Also, calculate the x- and y-components of the momentum.

Write the momentum vector of a 10 kg ball moving with velocity 40 m/s SE (-45 deg) in standard unit vector notation. Also, calculate the x- and y-components of the momentum.

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**Problem Statement:** Write the momentum vector of a 10 kg ball moving with a velocity of 40 m/s Southeast (-45 degrees) in standard unit vector notation. Also, calculate the x- and y-components of the momentum. **Solution:** 1. **Determine the Components of Velocity:** - The velocity vector can be split into x and y components using trigonometry. - Given angle: -45 degrees (Southeast direction). - Velocity in x-direction, \( v_x = v \cdot \cos(\theta) \) - \( v_x = 40 \, \text{m/s} \cdot \cos(-45^\circ) \) - Velocity in y-direction, \( v_y = v \cdot \sin(\theta) \) - \( v_y = 40 \, \text{m/s} \cdot \sin(-45^\circ) \) 2. **Calculate the Components:** - \( \cos(-45^\circ) = \sin(-45^\circ) = \frac{\sqrt{2}}{2} \) - \( v_x = 40 \, \text{m/s} \cdot \frac{\sqrt{2}}{2} = 20\sqrt{2} \, \text{m/s} \) - \( v_y = 40 \, \text{m/s} \cdot \frac{\sqrt{2}}{2} = -20\sqrt{2} \, \text{m/s} \) 3. **Calculate Momentum Components:** - Momentum, \( p = m \cdot v \) - Mass, \( m = 10 \, \text{kg} \) - Momentum in x-direction, \( p_x = m \cdot v_x \) - \( p_x = 10 \, \text{kg} \cdot 20\sqrt{2} \, \text{m/s} = 200\sqrt{2} \, \text{kg} \cdot \text{m/s} \) - Momentum in y-direction, \( p_y = m \cdot v_y \) - \( p_y = 10 \, \text{kg} \cdot (-20\sqrt{2}) \, \text{m/s} = -200
Transcribed Image Text:**Problem Statement:** Write the momentum vector of a 10 kg ball moving with a velocity of 40 m/s Southeast (-45 degrees) in standard unit vector notation. Also, calculate the x- and y-components of the momentum. **Solution:** 1. **Determine the Components of Velocity:** - The velocity vector can be split into x and y components using trigonometry. - Given angle: -45 degrees (Southeast direction). - Velocity in x-direction, \( v_x = v \cdot \cos(\theta) \) - \( v_x = 40 \, \text{m/s} \cdot \cos(-45^\circ) \) - Velocity in y-direction, \( v_y = v \cdot \sin(\theta) \) - \( v_y = 40 \, \text{m/s} \cdot \sin(-45^\circ) \) 2. **Calculate the Components:** - \( \cos(-45^\circ) = \sin(-45^\circ) = \frac{\sqrt{2}}{2} \) - \( v_x = 40 \, \text{m/s} \cdot \frac{\sqrt{2}}{2} = 20\sqrt{2} \, \text{m/s} \) - \( v_y = 40 \, \text{m/s} \cdot \frac{\sqrt{2}}{2} = -20\sqrt{2} \, \text{m/s} \) 3. **Calculate Momentum Components:** - Momentum, \( p = m \cdot v \) - Mass, \( m = 10 \, \text{kg} \) - Momentum in x-direction, \( p_x = m \cdot v_x \) - \( p_x = 10 \, \text{kg} \cdot 20\sqrt{2} \, \text{m/s} = 200\sqrt{2} \, \text{kg} \cdot \text{m/s} \) - Momentum in y-direction, \( p_y = m \cdot v_y \) - \( p_y = 10 \, \text{kg} \cdot (-20\sqrt{2}) \, \text{m/s} = -200
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