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velikii [3]
3 years ago
13

You throw a bouncy rubber ball and a wet lump of clay, both of mass m, at a wall. Both strike the wall at speed v, but while the

ball bounces off with no loss of speed, the clay sticks. What is the change in momentum of the clay and ball, respectively, assuming that toward the wall is the positive direction?
Physics
2 answers:
shutvik [7]3 years ago
8 0

Answer:

Change in momentum of the wet lump of clay = -mv

Change in momentum of the bouncy rubber ball = -2mv

Explanation:

Momentum, p = mv

Change in momentum of the wet lump of clay

The initial velocity of the wet lump of clay of mass, m is v, so its initial momentum is p₁ = mv. Since it sticks to the wall, its final velocity is zero. So its final momentum is p₂ = m × 0 = 0. Its change in momentum Δp = p₂ - p₁ = 0 - mv = -mv

Change in momentum of the bouncy rubber ball

The initial velocity of the rubber ball of mas m is v, so its initial momentum is p₃ = mv. Since it bounces of the wall with no loss of speed and thus moves in the opposite direction to its initial direction, its final velocity is -v. So its final momentum is p₄ = m × -v = -mv. Its change in momentum Δp = p₄ - p₃ = -mv - mv = -2mv

nadezda [96]3 years ago
6 0

Your answer


<h2>-mv; -2mv</h2>

Good luck!

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Answer:

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In the case of mass A,

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