Question

A marble rolls on the track shown in the figure below, with hs-22 cm and hc17 cm. If the marble has a speed of 2.8 m/s at point A, what is its speed at points B and C? It m/s m/s

Answer 1

consider the motion from A to B

v_A = speed at A = 2.8 m/s

v_B = speed at B = ?

h_A = height at A = 0 m

h_B = height at B = 22 cm = 0.22 m

m = mass of the marble

using conservation of energy between A and B

Total energy at A = Total energy at B

kinetic energy at A + potential energy at A = kinetic energy at B + potential energy at B

(0.5) m v_A² + m g h_A = (0.5) m v_B² + m g h_B

(0.5) v_A² + g h_A = (0.5) v_B² + g h_B

(0.5) (2.8)² + (9.8) (0) = (0.5) v_B² + (9.8) (0.22)

v_B = 1.88 m/s

consider the motion from A to C

v_A = speed at A = 2.8 m/s

v_C = speed at C = ?

h_A = height at A = 0 m

h_C = height at C = 17 cm = 0.17 m

m = mass of the marble

using conservation of energy between A and C

Total energy at A = Total energy at C

kinetic energy at A + potential energy at A = kinetic energy at C + potential energy at C

(0.5) m v_A² + m g h_A = (0.5) m v_C² + m g h_C

(0.5) v_A² + g h_A = (0.5) v_C² + g h_C

(0.5) (2.8)² + (9.8) (0) = (0.5) v_C² + (9.8) (0.17)

v_C = 2.12 m/s