A soccer player kicks a ball of mass 0.45 kg to the east - NSC Physical Sciences - Question 4 - 2019 - Paper 1

To find the velocity of the ball-container system after the collision, we apply the principle of conservation of momentum:

$ext{Total Initial Momentum} = ext{Total Final Momentum}$

Before the collision, the total momentum is: $p_{ ext{initial}} = m_{ ext{ball}} imes v_{ ext{ball}} + m_{ ext{container}} imes v_{ ext{container}}$

Substituting the known values: $p_{ ext{initial}} = (0.45 ext{ kg} imes 9 ext{ m·s}^{-1}) + (0.20 ext{ kg} imes 0 ext{ m·s}^{-1}) = 4.05 ext{ kg·m·s}^{-1}$

After the collision, the total momentum is: $p_{ ext{final}} = (m_{ ext{ball}} + m_{ ext{container}}) imes v_f$

Setting the two equal: $4.05 ext{ kg·m·s}^{-1} = (0.45 ext{ kg} + 0.20 ext{ kg}) imes v_f$

Solving for $v_f$ gives: v_f = rac{4.05 ext{ kg·m·s}^{-1}}{0.65 ext{ kg}} = 6.23 ext{ m·s}^{-1}

To determine the nature of the collision, we need to compare the kinetic energy before and after the collision:

Before collision: The total kinetic energy before the collision is calculated as:

KE_{ ext{initial}} = rac{1}{2} m_{ ext{ball}} v_{ ext{ball}}^2 = rac{1}{2} imes 0.45 ext{ kg} imes (9 ext{ m·s}^{-1})^2 = 18.225 ext{ J}

After collision: The total kinetic energy after the collision is:

KE_{ ext{final}} = rac{1}{2} (m_{ ext{ball}} + m_{ ext{container}}) v_f^2 = rac{1}{2} (0.45 ext{ kg} + 0.20 ext{ kg}) (6.23 ext{ m·s}^{-1})^2 KE_{ ext{final}} = rac{1}{2} (0.65 ext{ kg}) imes 38.8729 ext{ m}^2/ ext{s}^2 = 12.614 ext{ J}

Since the initial kinetic energy (18.225 J) is greater than the final kinetic energy (12.614 J), the collision is classified as inelastic.