Exercise \( 7.11 \mathrm{Q} 1(2012) \) (d)A proton travelling with a velocity of \( 3.00 \times 10^{7} \mathrm{~m} \mathrm{~s} \) collides with an oxygen nucleus, of mass \( 2.56 \times 10^{-26} \mathrm{~kg} \) that is at rest, and is scattered through an angle of 90 r . Calculate the velocity and direction of the oxygen nucleus using Newtonian mechanics. [12] (l)A rescue helicopter of mass 810 kg , supports itself in a stationary position by imparting a downward velocity, v , to the air in a circle of radius 4.0 m . The density of the air is 1.20 kg \( \mathrm{~m}^{-3} \). Calculate: (i)the value of v (ii)the power, P , required to support the helicopter [5]

To solve the problems presented, we will use the principles of conservation of momentum and the equations of fluid dynamics. ### Part (d): Collision of a Proton with an Oxygen Nucleus **Given:** - Mass of proton, \( m_p = 1.67 \times 10^{-27} \, \text{kg} \) - Initial velocity of proton, \( v_p = 3.00 \times 10^{7} \, \text{m/s} \) - Mass of oxygen nucleus, \( m_O = 2.56 \times 10^{-26} \, \text{kg} \) - Initial velocity of oxygen nucleus, \( v_O = 0 \, \text{m/s} \) - Angle of scattering of proton, \( \theta_p = 90^\circ \) **Conservation of Momentum:** In a two-body collision, the total momentum before the collision equals the total momentum after the collision. 1. **Before the collision:** - Total momentum in the x-direction: \[ p_{x, \text{initial}} = m_p v_p + m_O v_O = m_p v_p \] - Total momentum in the y-direction: \[ p_{y, \text{initial}} = 0 \] 2. **After the collision:** - Let the velocity of the proton after the collision be \( v_{p}' \) and the velocity of the oxygen nucleus be \( v_{O}' \). - The proton is scattered at \( 90^\circ \), so: \[ p_{x, \text{final}} = m_p v_{p}' \cos(90^\circ) + m_O v_{O}' \cos(\phi) \] \[ p_{y, \text{final}} = m_p v_{p}' \sin(90^\circ) + m_O v_{O}' \sin(\phi) \] Since \( \cos(90^\circ) = 0 \) and \( \sin(90^\circ) = 1 \): \[ p_{x, \text{final}} = m_O v_{O}' \cos(\phi) \] \[ p_{y, \text{final}} = m_p v_{p}' + m_O v_{O}' \sin(\phi) \] 3. **Setting up the equations:** - For the x-direction: \[ m_p v_p = m_O v_{O}' \cos(\phi) \quad (1) \] - For the y-direction: \[ 0 = m_p v_{p}' + m_O v_{O}' \sin(\phi) \quad (2) \] 4. **Assuming elastic collision:** In an elastic collision, the kinetic energy is also conserved. However, since we are only interested in the final velocities, we can solve the momentum equations directly. 5. **Solving for \( v_{O}' \):** From equation (1): \[ v_{O}' = \frac{m_p v_p}{m_O \cos(\phi)} \] Since the proton is scattered at \( 90^\circ \), we can assume \( \phi \) is the angle of the oxygen nucleus after the collision. For simplicity, we can assume it moves in the x-direction, so \( \cos(\phi) = 1 \). Thus: \[ v_{O}' = \frac{m_p v_p}{m_O} \] Substituting the values: \[ v_{O}' = \frac{(1.67 \times 10^{-27} \, \text{kg})(3.00 \times 10^{7} \, \text{m/s})}{2.56 \times 10^{-26} \, \text{kg}} \approx 1.96 \, \text{m/s} \] ### Part (l): Helicopter Supporting Itself **Given:** - Mass of helicopter, \( m = 810 \, \text{kg} \) - Radius of air circle, \( r = 4.0 \, \text{m} \) - Density of air, \( \rho = 1.20 \, \text{kg/m}^3 \) **(i) Calculate the value of \( v \):** The helicopter must create a downward velocity \( v \) in the air to support its weight. The force exerted by the air must equal the weight of the helicopter. 1. **Weight of the helicopter:** \[ F