Question

A lorry and a car with the same kinetic energy are brought to rest by the application of brakes which provide equal retarding forces. Which of them will come to rest in a shorter distance?

• A. Lorry
• B. Car
• C. Both will stop at the same distance
• D. Cannot be said

Answer 1

The correct option is C Both will stop at the same distance

Let us assume the two masses to be $m_1$ and $m_2$ having velocities $v_1$ and $v_2$ respectively. Since they both are acted upon by equal retarding forces, we can formulate $m_1{a}_1=m_2{a}_2$ where $a_{1}and{a}_2$ are their respective accelerations.
Therefore $a_1=\frac{m_2}{m_1}\times a_2$................($\star$)
It is given they have the same kinetic energies
$\Rightarrow \frac{m_1}{m_2}=\frac{(v_2{)}^2}{(v_1{)}^2}$..........(1)
Using the position-velociy equation which is $v^2-u^2=2as$ we get $s_1=\frac{(v_1{)}^2}{2a_1}$ .......(2) as it comes to rest, final velocity is zero. We neglect the sign and consider only magnitude in this case. Similarly , $s_2=\frac{(v_2{)}^2}{2a_2}$..............(3)
Now from eq.(1) we get $(v_1{)}^2=\frac{m_2}{m_1}\times (v_2{)}^2$.........(4)
Substitute the value of eq(4) and eq.($\star$) in eq. (2) we get
$\Rightarrow s_1=\frac{\frac{m_2}{m_1}\times (v_2{)}^2}{2\times \frac{m_2}{m_1}\times a_2}=s_2$
$s_1=s_2$ and hence both will stop after the same distance.