Answer: 3.288 m/s
Step-by-step explanation:
Given
Mass of Mary, m1 = 69 kg
Mass of Sue, m2 = 56 kg
Speed of Mary, v1 = 2.4 m/s
Speed of Sue, v2 = 4.4 m/s
Speed of the 2 of them, v = ?
We solve this using the principle of conservation of linear momentum
m1.v1 + m2.v2 = m1.v + m2.v
m1.v1 + m2.v2 = (m1 + m2) v
v = [(m1.v1) + (m2.v2)] / (m1 + m2)
v = [(69 * 2.4) + (56 * 4.4)] / (69 + 56)
v = (164.6 + 246.4) / 125
v = 411 / 125
v = 3.288 m/s
Thus, the speed at which both Mary and Sue slide together across the ice is 3.288 m/s