ICSE Class 9 Physics Question 8 of 20

Laws of Motion — Question 14

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Question 14

A car is moving with a uniform velocity of 30 m s-1. It is stopped in 2 s by applying a force of 1500 N through it's brakes. Calculate (a) the change in momentum of car, (b) the retardation produced in car, and (c) the mass of the car.

Answer

(i) We know that,

acceleration = vut\dfrac{\text v - \text u}{\text t}

Given,

v = 0

u = 30 m s-1

t = 2 s

f = 1500 N

Substituting the values in the formula above we get,

a=0302a=15 m s2\text a = \dfrac{0 - 30}{2} \\[0.5em] \Rightarrow \text a = - 15\ \text{m s}^{-2} \\[0.5em]

The negative sign shows that it is retardation. So retardation = 15 m s -2

Now,

Force (f) = mass (m) x acceleration (a)

Substituting the values in the formula above we get,

1500=m×(15)m=100 kg1500 = \text m \times (15) \\[0.5em] \Rightarrow \text m = 100 \text{ kg} \\[0.5em]

Hence,

Mass of the car = 100 kg

As we know,

Change in momentum = Final momentum - Initial momentum

Initial momentum = mu = 100 x 30 = 3000 kg m s-1

Final momentum = mv = 100 x 0 = 0

Substituting the values in the formula above we get,

Change in momentum = 0 - 3000 = - 3000 kg m s-1

Hence,

(i) change in momentum = 3000 kg m s-1

(ii) The retardation produced in the car = 15 m s -2

(iii) The mass of the car = 100 Kg