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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 4.80 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 68.0 kg and sue's is 55.0 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.)



Answer :

Mary and Sue would cross with a speed of 3.288 m/s

Given: m1 = 69 kg for Mary's mass

Sue weighs 56 kilograms per square meter.

Mary's v1 velocity is 2.4 m/s.

Sue's v2 speed is 4.4 m/s.

The two people's joint speed is v.

We resolve this using the linear momentum conservation concept.

The equation is:

m1.v1 + m2.v2 = m1.v + m2.v

m1.v1 + m2.v2 = (m1 + m2) v

= [(m1.v1) + (m2.v2)] / (m1 + m2) \sv

= [(69 * 2.4) + (56 * 4.4)] / (69 + 56) v

= (164.6 + 246.4) / 125 \s

= 411 / 125 \s

v = 3.288 m/s

Mary and Sue will therefore slide across the ice at a pace of 3.288 meters per second.

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