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.

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,

$s = \frac{u^2}{2a}$

We know,

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

The stopping distance becomes,

$s = \frac{u^2m}{2F}$

For object 1:

$s = \frac{(2)^2 X 4 }{F}$

$s = \frac{16}{F}$

For object 2:

$s = \frac{(4)^2 X 1}{F}\\ \\s = \frac{16}{F}$

For both the objects the distance travelled is same.

Therefore, both the objects travel equal distance before stopping.