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.

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

Answers

Answer 1

Answer:

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


Related Questions

A book of 500 leaves has a mass of 1kg if its thickness is 5cm what are the mass and thickness of each leaf

Answers

Answer:

0.002kg and 0.01cm

Explanation:

500 leaves has a thickness is 5cm

Means I leaf has a thickness of 5/500= 0.01cm

Similarly the mass of one leaf would be 1/500 =0.002kg

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