Question

What is the change in momentum of a car when 10,000 newton force is
applied in 10 seconds

Answer 1

AnswEr :

• Force (F) = 10000 N
• Time (t) = 10s

MomentuM : Momentum is product of mass and velocity. It's denoted by p.

By, 2nd law of Newton we have a equation that,

$\\ \longrightarrow \tt{\vec F\ =\ \dfrac{d \vec p}{dt}} \\ \\ \\ \longrightarrow \tt{d \vec p\ =\ \vec F.dt} \\ \\ \\ \longrightarrow \tt{d \vec p\ =\ 10000\ \times\ 10} \\ \\ \\ \longrightarrow \underline{\boxed{\frak{d \vec p\ =\ 100000\ kgms^{-1}}}}$

Answer 2

GIVEN :-

• Force , F = 10000 N.
• Time , t = 10 seconds.

TO FIND :-

• The change in momentum of a car , ∆p.

SOLUTION :-

As we know that the rate of change in momentum of any object is given by,

$\\ : \implies \displaystyle \sf \: \Delta p = m(v - u).$

Now, by using the second law of motion we have,

$: \implies \displaystyle \sf \: F = \frac{m(v - u)}{t}$

$: \implies \displaystyle \sf \:10000 \: kg.ms ^{ - 2} = \frac{m(v - u)}{10}$

$: \implies \displaystyle \sf \:10000 \: kg.ms ^{ - 2} \times 10s =m(v - u)$

$: \implies \displaystyle \sf \:100000 \: kg.ms ^{ - 2 + 1} =m(v - u)$

$: \implies \displaystyle \sf \:100000 \: kg.ms ^{ - 1} =m(v - u)$

$: \implies \displaystyle \sf \:100000 \: kg.ms ^{ - 1} =\Delta p$

$: \implies \underline{ \boxed{\displaystyle \sf \: change \: in \: momentum = 100000 \: kg.ms ^{ - 1}}}$