Question

Magnitude of impulse?

A 15.0g marble s dropped from rest onto the floor 1.44m below. If the marble bounces straight upward to a height of 0.640m, what are the magnitude and direction of the impulse delivered to the marble by the floor?

Answer

Let the velocity of strike is v1 m/s and the velocity after strike is v2 m/s

=>v1 = sqrt[2gh1] =sqrt[2 x 9.8 x 1.44] = 5.31 m/s

=>v2 = sqrt[2gh2] = sqrt[2 x 9.8 x 0.64] = - 3.54 m/s

delta v = v2 - v1 = -3.54 - 5.31 = -8.85 m/s

Delta P = m x delta v

=>Delta P = 15 x 10^-3 x (-8.85) = -0.13 N-s

By :-Impulse[I] (by marble) = delta P

=>I(marble) = -0.13 N-s

=>I(floor) = 0.13 N-s [+ve sign indicates the direction is downward]

Jun 18 at 0:54

impulse = change of momentum

When falling

mass = 15g = .015kg

g = 9.8 ms^-2

v = final velocity before hitting the ground,

u = initial velocity =0

v^2 = u^2 + 2gh

> v^2 = 2*9.8*1.44

> v^2 =28.224

> v = 5.31m/s

Momentum = mv = .7965 kg.m/s

When rising

v = final velocity at max height = 0

u = initial velocity after hitting the ground

v^2 = u^2 - 2gh

> 0= u^2 - 2*9.8*0.64

> u^2 = 12.544

> u = 3.541 m/s

momentum = mu = -.015*3.541 = -0.053 kg.m/s......negative because opposite direction

so change of momentum = mv - mu = .7965 +0.053 = 0.8495 kg.m/s

Jun 18 at 4:40