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Question: A motorcar of mass \({\text{ 1200 kg }}\)is moving along a straight line with a uniform velocity of ...

A motorcar of mass  1200 kg {\text{ 1200 kg }}is moving along a straight line with a uniform velocity of  90 km/h. {\text{ 90 km/h}}{\text{. }}its velocity is slowed down to  18 km/h {\text{ 18 km/h }}in  4 s {\text{ 4 s }}by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

Explanation

Solution

We have to calculate the force required to bring the velocity down from  90 km/h. {\text{ 90 km/h}}{\text{. }}to  18 km/h {\text{ 18 km/h }}in  4 s {\text{ 4 s }}. We know that force acting on a body is the product of its mass and acceleration. Mass is given and we have to find the acceleration. Using the initial velocity, final velocity and time find the acceleration and use that to find the force.
Formula used:
F=maF = ma(Where, F {\text{ F }}stands for the force acting on an object,  m {\text{ m }}stands for the mass of the object and  a {\text{ a }}stands for the acceleration of the object)
v=u+atv = u + at(Where,  v {\text{ v }}stands for the final velocity of the car,  u {\text{ u }}stands for the initial velocity of the car,  a {\text{ a }}stands for the acceleration of the car and  t {\text{ t }}stands for the time taken for the velocity change)

Step by step solution
The mass of the motorcar is given by,
m=1200 kgm = 1200{\text{ }}kg
The initial velocity is given by,
u=90 km/hu = 90{\text{ }}km/h
The final velocity is given by,
v=18 km/hv = 18{\text{ km/h}}
For calculation we have to convert the velocities into  m/s {\text{ m/s }}by multiplying with  518{\text{ }}\dfrac{5}{{18}}
Then we get,
u=90×518=25 m/su = 90 \times \dfrac{5}{{18}} = 25{\text{ m/s}}
And
v=18×518=5 m/sv = 18 \times \dfrac{5}{{18}} = 5{\text{ m/s}}
The time taken for the change in velocity is given by,
t=4st = 4s
We know that
v=u+atv = u + at
a=vuta = \dfrac{{v - u}}{t}
Substituting the values of  v,u {\text{ v,u }} and  t {\text{ t }} in the above equation, we get
a=5254=5 m/s2a = \dfrac{{5 - 25}}{4} = - 5{\text{ m/}}{{\text{s}}^2}
The negative sign shows that the acceleration is negative i.e. retardation.
The force is given by the product of mass and acceleration
F=maF = ma
Substituting the values of mass and acceleration in the expression for force, we get
F=1200×5=6000NF = 1200 \times - 5 = - 6000N
Therefore, the acceleration is  - 5 m/s2{\text{ - 5 m/}}{{\text{s}}^2}
The magnitude of force is:  - 6000 N{\text{ - 6000 N}}

Note Another method to solve this question is given below:
We know that Force is defined as the rate of change of momentum with time.
F=mvmutF = \dfrac{{mv - mu}}{t}
The initial momentum, mu=1200×25 = 30000 kg m/s{{ mu = 1200}} \times {\text{25 = 30000 kg m/s}}
The final momentum,  mv = 1200×5=6000 kg m/s{\text{ mv = 1200}} \times 5 = 6000{\text{ kg m/s}}
The time taken for the change in momentum, t = 4s {\text{ t = 4s }}
Substituting these values in the expression for force, we get
F=mvmut=6000300004F = \dfrac{{mv - mu}}{t} = \dfrac{{6000 - 30000}}{4}
F=240004=6000NF = \dfrac{{ - 24000}}{4} = - 6000N
We also know that
F=maF = ma
Form this equation, we get
a=Fma = \dfrac{F}{m}
a=60001200=5 m/s2a = \dfrac{{ - 6000}}{{1200}} = - 5{\text{ m/}}{{\text{s}}^2}
Again we get the same acceleration and force.
The acceleration is  - 5 m/s2{\text{ - 5 m/}}{{\text{s}}^2}
The magnitude of force is:  - 6000 N{\text{ - 6000 N}}