Question

Equal forces act on isolated bodies A and B. The mass of A is 3/5 times the mass of B. The magnitude of the acceleration of B is …….. times of A.

Answer 1

Explanation:

Let us consider the scenario with the help of the force formula. We know that,

F = ma

where, F = force,

m = mass

a acceleration

From this formula we can conclude that force is directly proportional to both mass and acceleration.

Given that the forces acting on both the isolated bodies A and B. So, let us assume the mass of body B be 'x'. Then the mass of body A will be 3/5x. Let the acceleration of the body B be y' and that of A be y.

Since the force of both the bodies are equal, we may equate them.

Putting the given value in the formula, we get:

F = ma

xy' = 3/5x y y' = 3/5y

Answer 2

$\frac{3}{5}$

Explanation:

$Given, \\ mass \: \: of \: \: A = mass \: \: of \: \: \frac{3B}{5} \\ we \: \: know \: \: that \\ \boxed{force \: = \: mass \times acceleration} \\A/Q, \\force \: \: of \: \: A \: = force \: \: of \: \: B \\ mass \: \: of \: \: A \times \: acceleration \: \: of \: \:A = mass \: \: of \: \: B \times acceleration \: \: of \: \: B \\ mass \: \: of \: \: \frac{3B}{5} \times acceleration \: \: of \: \:A = mass \: \: of \: \:B \times acceleration \: \: of \: \: B \\ \frac{3}{5} \times acceleration \: \: of \: \:A = acceleration \: \: of \: \:B \\ so \\ acceleration \: \: of \: \:B = \frac{3}{5} \times acceleration \: \: of \: \:A$