A smooth sphere A, of mass 5 kg, collides directly with another smooth sphere B, of mass 6 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2015

The initial kinetic energy (KE_initial) is: $KE_i = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2$ Substituting the given values: $KE_i = \frac{1}{2} (5)(7^2) + \frac{1}{2} (6)(1^2)$ $= \frac{1}{2} (5)(49) + \frac{1}{2} (6)(1)$ $= 122.5 + 3 = 125.5 ext{ J}$

The final kinetic energy (KE_final) after the collision is: $KE_f = \frac{1}{2} m_A v_1^2 + \frac{1}{2} m_B v_2^2$ Substituting the calculated final velocities: $KE_f = \frac{1}{2} (5)(3.18^2) + \frac{1}{2} (6)(4.18^2)$ Calculate: $KE_f = \frac{1}{2}(5)(10.1124) + \frac{1}{2}(6)(17.4724)$ $= 25.281 + 52.417 = 77.698 ext{ J}$

Loss in kinetic energy: $\text{Loss} = KE_i - KE_f = 125.5 - 77.698 = 47.802 ext{ J}$

Impulse can be calculated as the change in momentum: $I = \Delta p = m_A (v_1 - u_A)$ Substituting into the equation: $I = 5(3.18 - 7)$ $I = 5(-3.82) = -19.1 ext{ kg m s}^{-1}$

Taking the magnitude: $|I| = 19.1 ext{ kg m s}^{-1}$

Using the equation of motion: $v^2 = u^2 + 2gh$ Where:

• $u = 0$ (initial speed)
• $h = 3.2 m$
• $g = 9.81 m s^{-2}$ (acceleration due to gravity) Substituting values: $v^2 = 0 + 2(9.81)(3.2)$ $v^2 = 62.752$ $v = \sqrt{62.752} ≈ 7.93 ext{ m s}^{-1}$

Using the coefficient of restitution: $e = \frac{h}{H}$ Where:

• $H = 3.2 m$ (initial height) Substituting the coefficient of restitution: $\frac{1}{3} = \frac{h}{3.2}$ Thus, h = \frac{3.2}{3} = 1.067 m.$$

Final calculations show: $h ≈ 1.067 m$