Question

Two carts are sitting on air track (frictionless surface). Assume cart A has mass of .50 kg, cart B has mass of 1.0 kg. You push cart A and give it an initial speed of 5.0 m/s toward B which was originally at rest. The two carts stick together and move together. Right after collision, the air is turned off so the two carts slide to a stop due to kinetic friction. Assume the coefficient of kinetic friction is .20.

How far did the two carts slide on the track after collision before coming to a stop?

Answer 1

first we need to find final velocity of system after collision y using conservation of linear momentum

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

0.5 * 5 + 1* 0 = ( 0.5 + 1)* v

v = 1.667 m/s

using conservation of energy

0.5 M v^2 = u M gd

d = 0.5* 1.667^2 / (0.2* 9.8)

d = 0.709 m

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