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An egg and carrier have a combined mass of .1 kg. The egg and carrier hit the ground at a speed of 15 m/s and bounce up at a speed of 2 m/s. If the carrier is in contact with the ground for .023 seconds what is the force on the

carrier?

User Ygautomo
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1 Answer

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The change in velocity of the egg and carrier as they hit the ground is 15 m/s - 2 m/s = 13 m/s. The impulse (the change in momentum) experienced by the egg and carrier as they hit the ground is equal to the product of their mass and the change in velocity, which is 0.1 kg * 13 m/s = 1.3 kg m/s.

The average force experienced by the egg and carrier as they hit the ground is equal to the impulse divided by the time of contact with the ground, which is 1.3 kg m/s / 0.023 s = 56.5 N. This is the force on the carrier.

User RollerMobster
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