Name: Conservation of Mechanical Energy
Uploaded: 2019-07-19
Duration: 5 min 12 s
Description: When a body falls from a height towards the ground, its potential energy decreases but not the total mechanical energy. Let us understand this using data.

Question

A lorry and a car moving with the same momentum are brought to rest by the application of brakes, which provide equal retarding forces. Which of them will come to rest in a shorter time? Which will come to rest in less distance?

Accepted Answer

Let the mass of the lorry $= (m_{1})$

The mass of the car $= (m_{2})$

The initial momentum of the car and the lorry $= (p)$

The final momentum of the car and the lorry $= 0$

The retarding force on the car and the lorry $= (F)$

The time taken by the lorry to stop $= t_{1}$

The time taken by the car to stop $= t_{2}$

Using the second law of motion we can write for the lorry:

$\frac{p}{t_{1}} = F \Rightarrow t_{1} = \frac{p}{F} . . . . . (i)$

Similarly, for the car:

$t_{2} = \frac{p}{F} . . . . (i i)$

From(i) and(ii), it is clear that the car and lorry will take the same time to stop.

Let the distance covered by the lorry while coming to rest $= d_{1}$

And the distance covered by the car $= d_{2}$

Using work-energy theorem: For the lorry:

$F d_{1} = \frac{p^{2}}{2 m_{1}}$

$\Rightarrow d_{1} = \frac{p^{2}}{2 m_{1} F}$ .......(iii)

For the car:

$d_{2} = \frac{p^{2}}{2 m_{2}}$ ......(iv)

From(iii) and(iv) we get:

For $m_{1} > m_{2}$ , $d_{1} < d_{2}$ . Hence the lorry will stop after covering less distance.

A lorry and a car moving with the same momentum are brought to rest by the application of brakes, which provide equal retarding forces. Which of them will come to rest in a shorter time? Which will come to rest in less distance?

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