Question

a person of mass 30 kg pushes another person of mass 40 kg due to which the first man starts moving with an acceleration of 2.m/s2 the acceleration of the second person atbthat instant​

Answer 1

Given:

Mass of man A = 30 kg

Mass of man B = 40 kg

Acceleration of man A = 2m/s²

To find:

Acceleration of man B at that instant.

Solution:

Force exerted on man A = mass* acceleration

= 30*2 = 60N

According to Newton's third law, if man A exerts a force on man B then he should experience the same amount of force but in the opposite direction:

Force exerted on man B = 60 =N

60N = 40kg * a

Where a is the acceleration of mass B

a = 60/40 = 1.5m/s²

Therefore the acceleration of man B is 1.5 m/s²

Answer 2

Given info : a person of mass 30 kg pushes another person of mass 40 kg due to which the first man starts moving with an acceleration of 2.m/s².

To find : the acceleration of the second person at that instant.

solution : a/c to Newton's 3rd law of motion,“action and reaction are of equal in magnitudes but direct in opposite directions.”

here first person pushes with a force F to second person then to balance the force, second person reacts the same amount of force in just opposite direction.

here acceleration of first person is 2m/s²

so, force acts on first person by second person = mass of first person × acceleration

= 30kg × 2 m/s² = 60 N

here must be force exerts on the second person by first one = 60 N

⇒mass of second person × acceleration of second person = 60 N

⇒40 kg × a = 60 N

⇒a = 60/40 = 1.5 m/s²

Therefore the acceleration of second person must be 1.5 m/s² to balance the force.

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