Question

calculate the impulse necessary to stop a 1,500kg car travelling at 90 km h ^ -1


plz solve this ​

Answer 1

Sol : Here , M = 1, 500 kg

u = 90 km h^-1 = 25 ms ^-1

v = 0

• Impulse required to stop the car ,

I = Mv - Mu ( v-u ) = 1,500 ( 0-25 )

= - 37,500 Ns

The negative sign shows that the applied impulse brings about a retarding effect .

Answer 2

Explanation:

GIVEN :-

Mass of the car = 1500 kg

Initial speed of car = 1900 kmph

TO FIND :-

Impulse required to stop the moving car

SOLUTION :-

We have ,

m = 1500 kg

u = 90kmph = 25 m/s

v = 0 m/s

$large {\underline {\boxed {\bigstar {\red{ \sf{ \: 1kmph = \frac{5}{18} \: m {s}^{ - 1}}}}}}}$

$90kmph = 90 \times \frac{5}{18} {ms}^{ - 1}$

$90kmph \: = 25 {ms}^{ - 1}$

relation between momentum and

impulse is given by :-

$l = p_{1} - p_{2}$

⟹ ⟹

$\begin{gathered} \implies \sf \: I = m(v - u) \\ \\ \implies \sf \: I = 1500(0 - 25) \\ \\ \implies \sf \: I = 1500( - 25) \\ \\ \implies {\underline {\boxed {\blue{\sf {\: I = - 37500 \: kg.m {s}^{ - 1} }}}}}\end{gathered} </p><p>$

∴ The impulse required to stop the moving car of mass 1500kg moving with speed 90kmph is -37500 kgms⁻¹

ADDITIONAL INFO :-

◉ Impulse is a vector quantity and SI unit of impulse is dimensionall same as momentum i.e , Kgms⁻¹

◉ Impulse applied is equal to the change in momentum of a body