Question

Two cars $C_1$ and $C_2$ of masses $M_1$ and $M_2$ have similar tyres. Given that $M_1>M_2$ and initially both the cars are moving with the same speed. Let the minimum stopping distance for them be $x_1$ and $x_2$, then:

• A. $x_2>x_1$
• B. $x_1=x_2$
• C. $x_1>x_2$
• D. $noneofthese$

Answer 1

The correct option is B $x_1=x_2$

Given that,

Mass of $C_1$ car = $M_1$

Mass of $C_2$ car =$M_2$

Distance for first car = $x_1$

Distance for second car = $x_2$

Given condition is $M_1>M_2$

Now, the friction force is

$F=\mu mg$

Now, the retardation is

$ma=\mu mg$

$a=\mu g$

Now, the retardation is same for both so, time taken in stopping will also same

Now, from equation of motion for first car

$s=ut+\frac{1}{2}a{t}^2$

$x_1=\frac{1}{2}\times \mu g{t}^2$

Now for second car

$s=ut+\frac{1}{2}a{t}^2$

$x_2=\frac{1}{2}\times \mu g{t}^2$

Hence, the distance is same $x_1=x_2$