Question

Two cars of unequal masses use similar tyres. If they are moving with same initial speed, the minimum stopping distance:
A. Is smaller for the heavier car
B. Is smaller for the lighter car
C. Is same for both the cars
D. Depends on volume of the car

Answer 1

Hint: The cars will be stopping on applying brakes. So, when the tyres come in contact with the ground or the road, force of friction acts due to which the speed of the car decelerates or decreases and after moving for some distance it stops.

Formula Used:
Mathematically, the force of friction is written by the formula:
\[F = \mu mg\]……(i)
Where ‘m’ is the mass, ‘g’ is acceleration due to gravity and \[\mu \] is the coefficient of kinetic friction.

Complete step by step solution:
Given that the mass of both the cars is unequal. If they are moving with same, initial speed, when they will stop the force applied will be given by,
$F=ma$
Where ‘m’ is the mass and ‘a’ is the acceleration (In this case deceleration, as the speed of the cars is decreasing)
The above equation on rearranging can be written as,
\[a = \dfrac{F}{m}\]

Substituting the value of force from equation (i) in above equation we get
\[a = \dfrac{{\mu mg}}{m}\]
The equation becomes
\[a = \mu g\]
From the above equation it is clear that the minimum stopping distance for both the cars, depends only on coefficient of friction and acceleration due to gravity and is independent of masses. Therefore, the stopping distance for both the cars irrespective of their masses is the same.

Hence, Option C is the correct answer

Note: It is important to note that the force of friction will be equal to weight of the object if the two surfaces in contact are regular. On the other hand, if the surfaces are irregular, the force of friction will be greater than the weight of the object.