A ball of mass 0.1 kg bounces on a hard surface. Every time it hits the floor, it loses a quarter of its kinetic energy. If the ball is released from a height of 1.00 m, after how many bounces will the ball bounce no higher
than 0.25 m?
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The height to which the ball bounces is directly proportional to its kinetic energy. Therefore, if the ball loses a quarter of its kinetic energy with each bounce, it will also bounce to a quarter of its previous height with each bounce.
So, if the ball is initially released from a height of 1.00 m, after the first bounce it will reach a height of 0.75 m (since it loses a quarter of its kinetic energy), after the second bounce it will reach a height of 0.75^2 = 0.5625 m, after the third bounce it will reach a height of 0.75^3 = 0.421875 m, and so on.
We want to find the number of bounces after which the ball will bounce no higher than 0.25 m. This means we need to solve the equation 0.75^n = 0.25, where n is the number of bounces.
Taking the natural logarithm of both sides of the equation, we get n * ln(0.75) = ln(0.25). Solving for n, we get n = ln(0.25) / ln(0.75) ≈ 2.41.
Since the number of bounces must be an integer, we round up to the nearest whole number to get 3 bounces. Therefore, after 3 bounces, the ball will bounce no higher than 0.25 m.
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