Equal forces F act on isolated bodies A and B. The mass of B is 1/5 times that of A. The magnitude of the acceleration of A is:

Explanation

According to Newton's second law, F = ma, where F is the force, m is the mass, and a is the acceleration. Since equal forces F act on both bodies A and B, we can express their accelerations as follows:For body A: a_A = F/m_AFor body B: a_B = F/m_BGiven that the mass of B (m_B) is 1/5 times that of A (m_A), we can substitute:m_B = (1/5)m_ATherefore, the acceleration of B becomes: a_B = F/(1/5)m_A = 5F/m_ANow, we know that:a_A = F/m_AThus, by comparing the two accelerations, we find:a_A = (1/5) * a_BThis means that the magnitude of acceleration of A is indeed 1/5 times that of B.Options B, C, and D are incorrect because they either miscalculate the relationship between the accelerations based on the mass ratio or ignore the fundamental principle that acceleration is inversely proportional to mass when the same force is applied.