Question

A man standing on a trolley pushes another identical trolley (both trolleys are at rest on a rough surface), so that they are set in motion and stop after some time. If the ratio of mass of first trolley with man to mass of second trolley is 3, then find the ratio of the stopping distances of the second trolley to that of the first trolley. (Assume coefficient of friction to be the same for both the trolleys)

Answer 1

Given mass of trolley with man to mass of second trolley is 3. let mass of second trolley be m.

Since , the man on first trolley pushes the second trolley.

Momentum is same .

Let, $v_2$ be velocity of second trolley and $v_1$ be the first one
$p=m{v}_2=3m{v}_1$

$\Rightarrow v_1=\frac{p}{3m};v_2=\frac{p}{m}$

And since the frictional deceleration =$\mu g$ for both the trolleys.

We know that,
$v^2-u^2=2as$

$s_1=\frac{v_1^2}{2a}$

$s_2=\frac{v_2^2}{2a}$

$\Rightarrow \frac{s_2}{s_1}=\frac{\frac{p^2}{m^2}}{\frac{p^2}{9m^2}}=9$