Question

Standing on the edge of a tall cliff, a curious man drops a rock, [m_1], and [1.00 s] later, throws a second rock, [m_2], downwards with a speed of [19.6 m/s]. The mass of the second rock, [m_2], is twice the mass of [m_1]. Assume no friction from air. What is the mass of the first rock?

Answer 1

The mass of the first rock is required to solve the problem, but it cannot be calculated from the provided information because gravity affects all masses equally in a vacuum, and additional necessary information, like the height of the cliff or the time to impact, is not provided.

Step-by-step explanation:

The question involves the concepts of gravity and free fall under the influence of Earth's acceleration due to gravity. Since there is no air resistance, the two rocks will experience the same acceleration regardless of their mass. The mass of the first rock, m_1, is not directly calculable from the information provided because the mass does not affect the falling speed in a vacuum. However, it's implied in the question that knowing the mass of the second rock is somehow relevant, but without the distance of the fall or more information about how the masses are related beyond m_2 = 2m_1, we can't determine the mass m_1.

To further evaluate the problem, typically, one would look to kinematic equations or energy considerations; however, without additional data (height of the cliff or the time it takes for the rocks to hit the ground), we cannot determine m_1.

Answer 2

The mass of the first rock is 1 kg.

Step-by-step explanation:

To calculate the mass of the first rock, we can use the Principle of Conservation of Energy. Since both rocks fall from the same height and we assume no friction from air, the total mechanical energy of the system is conserved. The potential energy of the first rock at the edge of the cliff is converted to kinetic energy when it reaches the ground. The potential energy is given by m1gh, and the kinetic energy is given by 1/2m1v12. Since both rocks fall for the same amount of time, we can equate their kinetic energies: to 1/2m1v12 = 1/2m2v22. Since the mass of the second rock is twice the mass of the first rock, we can substitute 2m1 for m2. Solving for m1, we find m1 = (v2/v1)2 * 2m1.

Using the given values, v1 = 19.6 m/s and v2 = 0 m/s (since it was dropped), we can substitute into the equation to find m1. (0/19.6)2 * 2m1 = 1, so m1 = 9.8 m1. This means the mass of the first rock is 1 kg.