Question

A 4.0-kg object is moving with speed 2.0 m/s. A 1.0-kg object is moving with speed 4.0 m/s. Both objects encounter the same constant braking force, and are brought to rest. Which object travels the greater distance before stopping?

Answer 1

Both the objects travel equal distance before stopping.

Step-by-step 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,

`v^2 - u^2 = 2as\\\\`

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,

`s = (u^2)/(2a)`

We know,

`a = (F)/(m)`

The stopping distance becomes,

`s = (u^2m)/(2F)`

For object 1:

`s = ((2)^2 X 4 )/(F)`

`s = (16)/(F)`

For object 2:

`s = ((4)^2 X 1)/(F)\\ \\s = (16)/(F)`

For both the objects the distance travelled is same.

Therefore, both the objects travel equal distance before stopping.