Two boys of mass m1 = 40 kg, m2 = 50 kg are standing on skates on the ice. The first boy pushes off the other with a force

Two boys of mass m1 = 40 kg, m2 = 50 kg are standing on skates on the ice. The first boy pushes off the other with a force, the modulus of which is F1 = 10 H. Determine the modules of acceleration that the boys will receive when pushing off. Neglect the friction force.

m1 = 40 kg.
m2 = 50 kg.
F12 = 10 N.
a1 -?
a2 -?
According to Newton’s 3 laws, the force F12 with which the first boy acts on the second is equal in magnitude and oppositely directed to the force with which the second boy acts on the first F21: F12 = – F21.
The “-” sign indicates the opposite direction of the forces F12 and F21.
According to Newton’s 2 law, the acceleration of a body a is directly proportional to the force F and inversely proportional to the mass of the body m: a = F / m.
a1 = F21 / m1.
a1 = 10 N / 40 kg = 0.25 m / s2.
a2 = F12 / m2.
a2 = 10 N / 50 kg = 0.2 m / s2.
Answer: when repulsing, the boys will receive accelerations a1 = 0.25 m / s2 and a2 = 0.2 m / s2, directed in the opposite direction.