A person with a mass of 79.5 kg is initially at rest on the edge of a large stationary platform with a mass of 195 kg. The platform is supported by frictionless wheels on a horizontal surface. The person jumps off the platform, traveling a horizontal distance of 1.00 m while falling a vertical distance of 0.500 m to the ground. What is the final speed of the platform?

To find the final speed of the platform, we can use the principle of conservation of momentum. Initially, the total momentum of the system (person + platform) is zero since both are at rest.

When the person jumps off, they gain momentum in the horizontal direction while falling vertically. However, according to the conservation of momentum, the change in momentum of the person must be equal and opposite to the change in momentum of the platform.

To calculate the person's momentum change, we can use the equation p = mv, where p is momentum, m is mass, and v is velocity. The person's mass is 79.5 kg and their velocity is the final speed, which we'll denote as v_p.

The platform's mass is 195 kg, and its final velocity will be denoted as v_plat.

The momentum change of the person can be calculated using Δp_person = m_person * v_p - m_person * 0 (initial velocity).

Similarly, the momentum change of the platform can be calculated using Δp_platform = m_platform * v_plat - m_platform * 0 (initial velocity).

Since the momentum changes are equal and opposite, we can equate the two expressions:

m_person * v_p = -m_platform * v_plat

Simplifying this equation, we find:

v_plat = (-m_person / m_platform) * v_p

Plugging in the given values, we have:

v_plat = (-79.5 kg / 195 kg) * v_p

Simplifying further, we find:

v_plat = -0.4077 * v_p

Therefore, the final speed of the platform is 0.4077 times the final speed of the person.

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