Question

A car of mass 1000 kg moves uniformly at 10 m/s. If the engine of the car develops an extra linear momentum of 1000 kg m/s, calculate the new velocity with which the car runs.
$
  {\text{A}}{\text{. 22 m}}{{\text{s}}^{ - 1}} \\
  {\text{B}}{\text{. 11 m}}{{\text{s}}^{ - 2}} \\
  {\text{C}}{\text{. 55 m}}{{\text{s}}^{ - 1}} \\
  {\text{D}}{\text{. 11 m}}{{\text{s}}^{ - 1}} \\
$

Answer 1

Hint: The linear momentum of a body in motion is given as the product of the mass of the body and the linear velocity of the body. We can add the initial momentum to the increase in momentum to obtain new momentum from which the velocity of the car can be calculated easily.

Formula used:
The formula for linear momentum is given as
$p = m{\text{v }}...{\text{(i)}}$
where p represents the linear momentum of the body in motion, m represents the mass of the body while v is the velocity with which the body moves.

Complete answer:
All bodies which exhibit a linear motion possess a certain amount of linear momentum which is given as the product of the mass of the body and its velocity.
We are given that a car of a certain mass m initially has the velocity which is given as
$
  m = 1000kg \\
  {{\text{v}}_1} = 10m/s \\
$
Therefore, we can calculate the initial linear momentum of the car using the equation (i) as follows
${p_1} = m{{\text{v}}_1}$
Using the known values we get
$
  {p_1} = 1000 \times 10 \\
   = 10000kgm/s \\
$
Now it is given that the engine of the car increases the momentum of the car by an amount equal to 1000kgm/s. Therefore, the final linear momentum of the car can be written as
$
  {p_2} = {p_1} + 1000 \\
   = 10000 + 1000 \\
   = 11000kgm/s \\
$
Now we know the final linear momentum and mass of the car from which we can calculate the velocity of the car by using the same relation for linear momentum as follows:
${{\text{v}}_2} = \dfrac{{{p_2}}}{m}$
Using the known values in this equation we get
${{\text{v}}_2} = \dfrac{{11000}}{{1000}} = 11m/s$
This is the required answer so the correct option is option D.

Note: The student should be careful regarding the units that are being used in the options given. The units of velocity are m/s while $m/{s^2}$ are the units for acceleration through the option B which has the right magnitude, it is wrong.