Mary slides down a snow-covered hill on a large piece of cardboard and then slides across a frozen pond at a constant velocity of 2.40 m/s. After Mary has reached the bottom of the hill and is sliding across the ice, Sue runs after her at a velocity of 4.40 m/s and hops on the cardboard. How fast do the two of them slide across the ice together on the cardboard? Mary's mass is 69.0 kg and Sue's is 56.0 kg. Ignore the mass of the cardboard and any friction between the cardboard and the snow and/or ice. (Indicate the direction with the sign of your answer.) m/s

3.288 m/s

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

unbalanced force

explanation:

the speed of the cart after 3 seconds of low fan speed is ⇒ 54.0 cm/s.

the speed of the cart after 5 seconds of medium fan speed is ⇒ 120.0 cm/s.

the speed of the cart after 2 seconds of high fan speed is ⇒ 64.0 cm/s. 

just did the lab.