Question

Calculate the change in momentum of a car of mass $1000kg$ when it’s speed increases from $36km{h^{ - 1}}$ to $108km{h^{ - 1}}$ uniformly?

Answer 1

Hint: In this question, we are asked to find the change in the momentum occurring as the speed of the car increases uniformly. The momentum of a body of mass \[m\]and moving with speed/velocity$v$ is given as the product of mass and speed/velocity of the body. Momentum can be found out at a particular moment if you know the velocity of the moving body at that particular moment. The difference between the values of momentum will give you the change in the momentum.

Complete step by step answer:
Mathematically, the momentum is given as $mv$ and is denoted by $p$. Therefore $\Delta P = \Delta (mv)$.
The initial speed of the car is ${v_i} = 36$$km{h^{ - 1}}$$ = 36 \times 1000 \div (60 \times 60) = 36 \times 5 \div 18 = 10$$m{s^{ - 1}}$ and the final speed of the car is ${v_f} = 108$$km{h^{ - 1}}$$ = 108 \times 5 \div 18 = 30$$m{s^{ - 1}}$
The initial momentum will be ${p_i} = m{v_i}$ and the final momentum will be ${p_f} = m{v_f}$
The change in momentum will be
$
  \Delta P = \Delta (mv) \\
\Rightarrow \Delta P = m({v_f} - {v_i}) \\
\Rightarrow \Delta P = 1000(30 - 10) \\
\Rightarrow \Delta P = 20000 \\
 $
$\therefore \Delta P = 20000$$kgm{s^{ - 1}}$
Therefore, the change in momentum of a car of mass $1000kg$ when it’s speed increases from $36km{h^{ - 1}}$ to $108km{h^{ - 1}}$ uniformly is $20000$ $kgm{s^{ - 1}}$.
Additional information: The second law of motion says that the force on a body of mass $m$ moving with an acceleration equal to $a$ is given as $F = ma$.
The actual expression for force is given as $F = \dfrac{{dp}}{{dt}}$. The force is equal to the rate of change of momentum with respect to time.
$
  F = \dfrac{{dp}}{{dt}} \\
\Rightarrow p = mv \\
\Rightarrow F = \dfrac{{d(mv)}}{{dt}} \\
\Rightarrow F = \dfrac{{dm}}{{dt}}v + m\dfrac{{dv}}{{dt}} \\
 $
*Product rule is applied to find the derivative.
Usually we deal with cases where mass is constant, so the 1st term of the force becomes zero and we are left with $F = m\dfrac{{dv}}{{dt}}$.

Note:
Here we have found the change of momentum, without thinking about whether it is increasing uniformly or not. Actually, it does not matter how the momentum changes, it only depends on the initial and final value of the momentum. Because no matter what path you take to reach from the initial state to final state, the end result will be the same. Keep in mind that the above discussion is valid only if the infinitesimal change in the momentum is expressed explicitly and not implicitly, i.e. $dp = f$ where $f$ is any function not involving $dp$.