Sphere A of mass 400 g is travelling horizontally with a speed of 6 m s⁻¹ when it collides with sphere B of mass 150 g travelling in the opposite direction with a speed of 9 m s⁻¹ - Leaving Cert Physics - Question b - 2017

Using the conservation of momentum, we have the equation:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

• $m_1$ = mass of sphere A = 0.4 kg (400 g)
• $u_1$ = initial velocity of sphere A = 6 m s⁻¹
• $m_2$ = mass of sphere B = 0.15 kg (150 g)
• $u_2$ = initial velocity of sphere B = -9 m s⁻¹ (negative due to opposite direction)
• $v_1$ = final velocity of sphere A = 0 m s⁻¹ (comes to rest)
• $v_2$ = final velocity of sphere B (to be calculated)

Substituting the known values:

$(0.4)(6) + (0.15)(-9) = (0.4)(0) + (0.15)v_2$