Question

Two cars of unequal masses use similar tyres. If they are moving with the same initial speed the minimum slipping distance
(a) is smaller for heavier car
(b) is smaller for lighter car
(c) is same for both cars
(d) depends on the volume of the car

Answer 1

Answer ⇒ Option (c). The minimum stopping distance is same for both the cars.

Explanation ⇒  Let us take an help of the mathematical expression to derive our result.

We know that when car will be stopped then the Force of Friction acting will be the Limiting Friction.

∴ f = μmg

Now, ma =  μmg

⇒ a =  μg

Since, the acceleration is independent of the mass thus the minimum stopping distance will also be independent of the mass as they have the same initial speed. [Taking help of the Second Equation of motion.]

∴ Option (c). is correct answer.

Hope it helps.

Answer 2

Given that,

Two cars of unequal masses use similar tyres.

To find,

The minimum slipping distance if they are moving with the same initial speed.

Solution,

Let the masses of two cars are m₁ and m₂, where m₁>m₂

For slipping,

Force of friction for the first car, F₁=μm₁g, μ = coefficient of friction.

The force of friction for the second car, F₂=μm₂g

Now using equation of motion :

$v^2-u^2=2as$

Let v = 0 finally it stops

$s=\dfrac{u^2}{2a}$.....(1) , a is acceleration of car

The acceleration for car 1, $a_1=\dfrac{F}{m_1}=\dfrac{\mu m_1g}{m_1}=\mu g$

The acceleration for car 2, $a_2=\dfrac{F}{m_2}=\dfrac{\mu m_2g}{m_2}=\mu g$

The acceleration of car 1 = car 2

So, from equation (1), we find that the stopping distance for both cars is same. Hence, the correct option is (c).