A smooth sphere A, of mass 2 kg, collides directly with another smooth sphere B, of mass 3 kg, on a smooth horizontal table - Leaving Cert Applied Maths - Question 5 - 2007

To calculate the loss in kinetic energy, we need to find the kinetic energy before and after the collision:

1. Kinetic Energy before collision:
   The formula for kinetic energy is:

   $KE = \frac{1}{2} mv^2$

   So,

   • For sphere A:

   $KE_A = \frac{1}{2}(2)(5^2) = \frac{1}{2}(2)(25) = 25 ext{ J}$

   • For sphere B:

   $KE_B = \frac{1}{2}(3)(2^2) = \frac{1}{2}(3)(4) = 6 ext{ J}$

   • Total KE before collision:

   $KE_{total ext{ before}} = KE_A + KE_B = 25 + 6 = 31 ext{ J}$

2. Kinetic Energy after collision:
   Using the speeds from part (i):

   • For sphere A:

   $KE'_A = \frac{1}{2}(2)(2^2) = \frac{1}{2}(2)(4) = 4 ext{ J}$

   • For sphere B:

   $KE'_B = \frac{1}{2}(3)(4^2) = \frac{1}{2}(3)(16) = 24 ext{ J}$

   • Total KE after collision:

   $KE_{total ext{ after}} = KE'_A + KE'_B = 4 + 24 = 28 ext{ J}$

3. Loss in Kinetic Energy:

   = 31 - 28 = 3 ext{ J}$$

The loss in kinetic energy due to the collision is 3 J.

Impulse is defined as the change in momentum of an object. The formula is:

$Impulse = \Delta p = m(v_B - u_B)$

Where:

• $m = 3 ext{ kg}$ (mass of sphere B)
• $v_B = 4 ext{ m/s}$ (final velocity of B)
• $u_B = 2 ext{ m/s}$ (initial velocity of B)

Calculating the impulse:

$Impulse = 3(4 - 2) = 3(2) = 6 ext{ Ns}$

Thus, the magnitude of the impulse imparted to B due to the collision is 6 Ns.