Both the objects travel equal distance before stopping.
Explanation:
Given-
Mass of object 1, m₁ = 4kg
Speed of object 1, v₁ = 2m/s
Mass of object 2, m₂ = 1kg
Speed of object 2, v₂ = 4m/s
Force₁ = Force₂ = F
Distance, s = ?
We know,
[tex]v^2 - u^2 = 2as\\\\[/tex]
Where, v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
When the brake is applied, the object comes to rest and the final velocity, v becomes 0. So,
[tex]s = \frac{u^2}{2a}[/tex]
We know,
[tex]a = \frac{F}{m}[/tex]
The stopping distance becomes,
[tex]s = \frac{u^2m}{2F}[/tex]
For object 1:
[tex]s = \frac{(2)^2 X 4 }{F}[/tex]
[tex]s = \frac{16}{F}[/tex]
For object 2:
[tex]s = \frac{(4)^2 X 1}{F}\\ \\s = \frac{16}{F}[/tex]
For both the objects the distance travelled is same.
Therefore, both the objects travel equal distance before stopping.
