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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 0150 kg lump of clay is dropped from a height of 145 m onto the floor It sticks to the floor and does not bounce What is the magnitude of the impulse imparted class=

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MathPhys

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

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