Question

If the marble bounces straight upward to a height of 0.64 m, what is the magnitude of the impulse delivered to the marble by the floor?

Answer 1

0.133kg.m/s

Step-by-step explanation:

Using equation of motion

v² = u² + 2as where v is the final velocity in m/s, a = g = 9.81 m/s² and s is the height and u = 0

v² = 2×9.81×1.44m

v = √(2×9..81×1.44) = 5.312 m/s taking downward movement as negative

v = -5.312 m/s

the marble bounces straight upward to a height of 0.64 m

again

v² = u² + 2as where final velocity = 0

-u² = (2 × -9.81 × 0.64)

u = √12.5568 = 3.544m/s

impulse = m × Δv = 0.15 × ( 3.544 --5.312 ) = 0.15 × (3.544+5.312) = 0.133kg.m/s