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3 years ago

7

At an outdoor market, a bunch of bananas is set on a spring scale to measure the weight. The spring sets the full bunch of banan

as into vertical oscillatory motion, which is harmonic with an amplitude 0.14 m. The maximum speed of the bananas is observed to be 2 m/s. What is the mass of the bananas

1 answer:

olga55 [171]3 years ago

5 0

Answer:

Mass of banana is $1.12$ Kg

Explanation:

Step 1: Determine the equation of speed of an object moving in an harmonic motion

Speed of moving in an harmonic motion is given by

$v = \sqrt{\frac{k}{m} (A^2 -x^2)} \\$

Here, v represents the speed of the object in harmonic motion, k is the springs constant, m is the mass of the object, A is the amplitude, and x is the position.

In this question , $x = 0$ because only at this position maximum speed occurs

So the simplified equation becomes -

$v = \sqrt{(\frac{k}{m} * A)}$

OR

$m = \frac{kA^2}{v_(max)^2}$

Substituting the given values in above equation we get -

Assume spring constant is $16$ N/m

$m = \frac{16 * 0.14}{2} \\m = 1.12$

Mass of banana is $1.12$ Kg

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4 years ago

If two objects A and B have the same kinetic energy but A has three times the momentum of B, what is the ratio of their inertias

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- The momentum of an object is given by $p=mv$

- The inertia of an object is proportional to its mass, so we can write $I=km$ , where k just indicates a constant of proportionality

In this problem, we have:

- $K_A = K_B$ (the two objects have same kinetic energy)

- $p_A = 3 p_B$ (A has three times the momentum of B)

Re-writing both equation we have:

$\frac{1}{2}m_A v_A^2 = \frac{1}{2}m_B v_B^2\\m_A v_A = 3 m_B v_B$

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So, B has 9 times more mass than A, and so B has 9 times more inertia than A, and their ratio is:

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