A smooth sphere A, of mass 2 kg, collides directly with another smooth sphere B, of mass 5 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2014

Using the conservation of momentum before and after the collision for both B and A:

$m_A v_{1i} + m_B v_{2i} = m_A v_{1f} + m_B v_{2f}$

Where:

• $m_A = 2$ kg, $v_{1i} = 4$ m s⁻¹ (initial speed of A)
• $v_{2f} = 3$ m s⁻¹ (final speed of B)

Plugging in the values:

$2 imes 4 + 5 imes 2 = 2 v_{1f} + 5 imes 3$

This simplifies to: $18 = 2 v_{1f} + 15$ $3 = 2 v_{1f}$

Thus: $v_{1f} = 1.5 ext{ m s}^{-1}$

The coefficient of restitution (e) is defined as the ratio of the relative speeds after and before collision:

$e = \frac{v_{2f} - v_{1f}}{v_{1i} - v_{2i}}$

Using the speeds we found:

• $v_{2f} = 3$ m s⁻¹
• $v_{1f} = 1.5$ m s⁻¹
• $v_{1i} = 4$ m s⁻¹
• $v_{2i} = 2$ m s⁻¹

Substituting these values gives:

$e = \frac{3 - 1.5}{4 - 2} = \frac{1.5}{2} = \frac{3}{4}$

We can calculate the initial and final kinetic energies:

Initial K.E.: $KE_{initial} = \frac{1}{2} m_A v_{1i}^2 + \frac{1}{2} m_B v_{2i}^2$ Substituting: $KE_{initial} = \frac{1}{2} (2)(4^2) + \frac{1}{2} (5)(2^2) = 16 + 10 = 26$

Final K.E.: $KE_{final} = \frac{1}{2} m_A v_{1f}^2 + \frac{1}{2} m_B v_{2f}^2$ Substituting: $KE_{final} = \frac{1}{2} (2)(1.5^2) + \frac{1}{2} (5)(3^2) = 2.25 + 22.5 = 24.75$

Thus, the loss in kinetic energy: $KE_{loss} = KE_{initial} - KE_{final} = 26 - 24.75 = 1.25 ext{ J}$