Question

a force produces an acceleration of 4m/s^2 in a body of mass m1 and the same force produces an acceleration of 6 m/s^2 in another body ofass m2 If the same force is applied to ( m1 + m2 ) then acceleration will be :

Answer 1

We know that the force applied separately on both the masses is equal. The force on first body of mass m1, F= ma = 4m1. And on the second body, F= ma= 6m2. Now, 4m1=6m2 so, m2= 2/3m1. Now, force on second body was 6m2 so, 6m2= 6* (2/3m1) = 4m1. Now, when the same force is applied on the two bodies together, Mass= m1+m2= m1 + 2/3m1 = 5/3m1 and Force= 4m1 Using a= F/m, a= 4m1/(5/3m1) = 2.4 m/s^2.

Answer 2

Answer:

The acceleration will be 2.4 m/s².

Explanation:

Given that,

Mass of first body = m₁

Mass of second body = m₂

Acceleration of first body = 4 m/s²

Acceleration of second body = 6 m/s²

If the force is same on first and second body.

The force on first body

$F=4m_{1}$....(I)

The force on second body

$F=6m_{2}$.....(I)

Both forces are same .

So, The mass of second body is

$4m_{1}=6m_{2}$

$m_{2}=\dfrac{2}{3}m_{1}$

The force is

Put the value of m₂ in equation (II)

$F=6\times\dfrac{2}{3}m_{1}$

$F = 4m_{1}$

Now, The same force is applied to (m₁+m₂) then the acceleration will be

Using newton's formula

$F = ma$

$a = \dfrac{F}{m}$

$a =\dfrac{4m_{1}}{m_{1}+m_{2}}$....(III)

Put the value of m₂ in equation (III)

$a =\dfrac{4m_{1}}{m_{1}+\dfrac{2}{3}m_{1}}$

$a = 2.4\ m/s^2$

Hence, The acceleration will be 2.4 m/s².