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Two objects with masses M and m (M > m) are on a frictionless surface. A force F will accelerate the smaller object with an acceleration a. If that same force is applied to the larger object then it will (A) move with a greater acceleration. (B) move with the same acceleration. (C) move but with a smaller acceleration. (D) move only if the force F is greater than some minimum value.

Two objects with masses M and m (M > m) are on a frictionless surface. A force F will accelerate the smaller object with an acceleration a. If that same force is applied to the larger object then it will
(A) move with a greater acceleration.
(B) move with the same acceleration.
(C) move but with a smaller acceleration.
(D) move only if the force F is greater than some minimum value.

Grade:upto college level

2 Answers

Navjyot Kalra
askIITians Faculty 654 Points
8 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.

HRpatel
15 Points
3 years ago
The correct answer is (C) move but with a smaller acceleration.
A force acting on object 1 is F and
A force acting on object 2 is F'.
Here, we applied same force on both objects.
So F' = F.
   F = ma --- (1)
   F = Ma'  --- (2)
From (1) and (2) we get, 
 M = F/a  &   m = F/a'
Here we have, 
M > m
So, 
F/a' > F/a
a > a'
 

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