# Selina Solutions Class 9 Physics Solutions for Exercise 3C in Chapter 3 - Chapter 3 Laws Of Motion

Question 32 Exercise 3C

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

Given: t=2s, F=1500N, u=30m/s, v=0m/s,

acceleration = (v-u)/t

= (0-30)/2 = -15 m/s^2

The negative acceleration just indicates retardation.

Force = mass x acceleration

→ Mass = force/acceleration

→ = 1500/(15)

= 100kg

(a) To find change in momentum

Change in momentum = final velocity – initial velocity or Δp = m(v-u)

Δp = 100(0-30)

= 3000 kg m/s

(b) Retardation produced in car

We know that retardation is negative acceleration

∴acceleration = (v-u)/t

= (0-30)/2 = -15 m/s^2

Retardation is 15 m/s^2

(c) Mass of the car

As per newton’s second law of motion,

Force = mass x acceleration

→ Mass = Force/acceleration

→ = 1500/15

→ = 100kg

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