A 0.150 kg lump of clay is dropped from a height of 1.45 m onto the floor. It sticks to the floor and does not bounce. What is the magnitude of the impulse imparted to the clay by the floor during the impact? Assume that the acceleration due to gravity is =9.81 m/s2.

J = 0.800 kg m/s

Fmax = 291 N

Explanation:

During the fall, energy is conserved.

PE = KE

mgh = ½ mv²

v = √(2gh)

v = √(2 × 9.81 m/s² × 1.45 m)

v = 5.33 m/s

Alternatively, you can use kinematics to find the velocity.

Impulse = change in momentum

J = Δp

J = mΔv

J = (0.150 kg) (5.33 m/s)

J = 0.800 kg m/s

Impulse = area under F vs t graph

J = ∫ F dt

J = ½ Fmax Δt

(0.800 kg m/s) = ½ Fmax (0.0055 ms)

Fmax = 291 N

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )