- Chapter 1 – Measurements and Experimentation
- Chapter 2 Motion In One Dimension
- Chapter 3 Laws Of Motion
- Chapter 4 Pressure In Fluids And Atmospheric Pressure
- Chapter 5 Upthrust In Fluids Archimedes Principle And Floatation
- Chapter 6 Heat And Energy
- Chapter 7 Reflection Of Light
- Chapter 8 Propogation Of Sound Waves
- Chapter 9 Current Electricity
- Chapter 10 Magnetism
Selina Solutions Class 9 Physics Solutions for Exercise 3C in Chapter 3 - Chapter 3 Laws Of Motion
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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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