Navjyot Kalra
Last Activity: 9 Years ago
The correct option is (C) move but with a smaller acceleration.
Force acting on an object is equal to the mass of the object time’s acceleration of the object.
So, Force acting on an object (F) having mass “m” and acceleration “a” will be,
F = ma …… (1)
And
Force acting on an another object (Fʹ) having mass “M” (M>m) and acceleration “aʹ” will be,
Fʹ = Maʹ …… (2)
Divining equation (1) by (2), we get,
F/ Fʹ = ma/ Maʹ …… (3)
Since both the forces acting on both the different masses is same (F = Fʹ), therefore from equation (3) we get,
ma/ Maʹ = 1
a/aʹ = M/m …… (4)
Since M is heavier than m (M>m), therefore,
a/aʹ >1
So, aʹ < a …… (5)
Therefore larger object will move but with a smaller acceleration.
Thus from the above observation we conclude that, option (C) is correct.