Class 9 Physics Chapter 2 numerical problems are according to the new syllabus of the Punjab Board. Chapter 2 is related to Kinematics which is the branch of Physics that deals with the study of motion without discussing the effects.

## Class 9 Physics Chapter 2 Numerical Problems

The problems have been solved with easy-to-use methods using the following formulae:

S = v x t

v_{f} = v_{i} + at

S = v_{i}t + 1/2 at^{2}

2aS = v_{f}^{2} – v_{i}^{2}

V_{av} = (v_{f} + v_{i})/2

### Numerical Problems Chapter 2 – Kinematics

**2.1 A train moves with a uniform velocity of 36 kmh-1 for 10 s. Find the distance travelled by it. (100 m)**

**Solution: **

Velocity = v = 36 kmh^{-1} = 36 x 1000/ 60 x 60 = 36000/ 3600 = 10 ms^{-1}

Time t = 10 s

Distance = S =?

S = v x t

S = 10 x 10

S = 100 m

**2.2 A train starts from rest. It moves through 1 km in 100 s with uniform acceleration. What will be its speed at the end of 100 s.? (20 ms**^{-1})

^{-1})

**Solution: **

Initial velocity V_{i }= 0 ms^{-1}

Distance S = 1 km = 1000 m

Time = 100 s

Final velocity V_{f} =?

S = V_{i} x t +1/2 x a x t2

1000 = 0 x 100 + 1/2 x a x (100)^{2}

1000 = 1/2 x 10000a

1000 = 5000a

a = 1000/5000 = 0.2 ms^{-2}

Now using 1^{st} equation of motion

V_{f} = V_{i} + at

V_{f} = 0 + 0.2 x 100

V_{f} = 20 ms^{-1}

**2.3 A car has a velocity of 10 ms**^{-1}. It accelerates at 0.2 ms^{-2} for half minute. Find the distance travelled during this time and the final velocity of the car. (390 m, 16 ms^{-1})

^{-1}. It accelerates at 0.2 ms

^{-2}for half minute. Find the distance travelled during this time and the final velocity of the car. (390 m, 16 ms

^{-1})

**Solution: **

Initial velocity = V_{i = }10 ms^{-1}

Acceleration a = 0.2 ms^{-2}

Time t = 0.5 min = 0.5 x 60 = 30 s

Distance S =?

Final velocity V_{f} =?

S = V_{i }x t + 1/2 x a x t^{2}

S= 10 x 30 + 1/2 x 0.2 x (30)^{2}

S = 300 + 1/2 x 0.2 x 90

S = 300 + 1/2 x 2/10 x 90

S = 300 + 90

S = 390 m

Using 1^{st} equation of motion

V_{f} = V_{i} + at

V_{f }= 10 + 0.2 x 30

V_{f} = 10 + 6

V_{f} = 16 ms^{-1}

**2.4 A tennis ball is hit vertically upward with a velocity of 30 ms**^{-1}, it takes 3 s to reach the highest point. Calculate the maximum height reached by the ball. How long it will take to return to the ground? (45 m, 6 s)

^{-1}, it takes 3 s to reach the highest point. Calculate the maximum height reached by the ball. How long it will take to return to the ground? (45 m, 6 s)

**Solution:**

Initial velocity = V_{i = }30 ms^{-1}

Acceleration due to gravity g = -10 ms^{-2}

Time to reach maximum height = t = 3 s

Final velocity v_{f }= 0 ms^{-1}

Maximum height attained by the ball S =?

Time taken to return to ground t =?

S = V_{i }x t + 1/2 x g x t^{2}

S = 30 x 3 + 1/2 x (-10) x (3)^{2}

S = 90 – 5 x 9

S = 90 – 45

S = 45 m

Total time = time to reach maximum height + time to return to the ground

t = 3 s + 3 s

t = 6 s

**2.5 A car moves with a uniform velocity of 40 ms**^{-1} for 5 s. It comes to rest in the next 10 s with uniform deceleration. **Find** **deceleration** and **total distance travelled by car.**

^{-1}for 5 s. It comes to rest in the next 10 s with uniform deceleration.

**(-4 ms**^{-2}, 400 m)

^{-2}, 400 m)

**Solution:**

Initial velocity = Vi = 40 ms^{-1}

Time = t = 5 s

Final velocity = Vf = 0 ms^{-1}

Time = 10 s

deceleration a =?

total distance S =?

V_{f} = V_{i} + at

Or

at = V_{f – }V_{i}

a = (V_{f – }V_{i})/t

a = 0 – 40/10

a = -4 ms^{-2}

Total distance travelled = S = S_{1} + S_{2}

By using this relation

S_{1} = v x t

S_{1} = 40 x 5

S_{1} = 200 m ………………………………. (i)

Now by using 3^{rd} equation of motion

2aS_{2} = V_{f}^{2} – V_{i}^{2}

S_{2} = (V_{f}^{2} – V_{i}^{2})/2a

S_{2} = [(0)2 – (40)2]/2 x (-4)

S_{2} = -1600/-8

S_{2} = 200 m ……………………………………… (ii)

From (i) and (ii) we get;

S = S_{1} + S_{2}

Or

S = 200 m + 200 m

S = 400 m

**2.6 A train starts from rest with an acceleration of 0.5 ms**^{-2}. Find its speed in kmh-1, when it has moved through 100 m. (36 kmh^{-1})

^{-2}. Find its speed in kmh-1, when it has moved through 100 m. (36 kmh

^{-1})

**Solution:**

Initial velocity V_{i} = 0 ms^{-1}

Acceleration a = 0.5 ms^{-2}

Distance S = 100 m

Final velocity V_{f} =?

2aS = Vf^{2} – V_{i}^{2}

2 x 0.5 x 100 = Vf^{2} – 0

Or

100 = Vf^{2}

Or

Vf^{2 }= 100 ms^{-1}……………………..(I)

Speed in kmh^{-1}:

From (I) we get;

v_{f }= 10 x 3600/1000

v_{f} = 36 kmh^{-1}

**2.7 A train starting from rest, accelerates uniformly and attains a velocity of 48 kmh**^{-1} in 2 minutes. It travels at this speed for 5 minutes. Finally, it moves with uniform retardation and is stopped after 3 minutes. Find the total distance travelled by train. (6000m)

^{-1}in 2 minutes. It travels at this speed for 5 minutes. Finally, it moves with uniform retardation and is stopped after 3 minutes. Find the total distance travelled by train.

**Solution: **

**Case – I: **

Initial velocity = V_{i} = 0ms^{-1}

Time = t = 2 minutes = 2x 60 = 120 s

Final velocity = V_{f} = 48 kmh^{-1} = 48 x 1000/3600 = 13.333 ms^{-1}

S_{1} = V_{av} x t

S_{1} = (V_{f} + V_{i})/2 x t

S_{1} = 13.333 + 0/2 x 120

S_{1} = 6.6665 x 120

S_{1} = 799.99 m = 800 m

**Case – II:**

Uniform velocity = V_{f} = 13.333 ms^{-1}

Time = t = 5 minutes = 5 x 60 = 300 s

S_{2} = v x t

S_{2} = 13.333 x 300

S_{2} = 3999.9 = 4000 m

**Case – III:**

Initial velocity = V_{f }= 13.333 ms^{-1}

Final velocity = V_{i} = 0 ms^{-1}

Time = t = 3 minutes = 3 x 60 = 180 s

S_{3} = V_{av} x t

S_{3} = (V_{f }+ V_{i})/2 x 180

S_{3} = 13.333 + 0/2 x 180

S_{3} = 6.6665 x 180

S_{3} = 1199.97 = 1200 m

Total distance = S = S_{1} + S_{2 }+ S_{3}

S = 800 + 4000 + 1200

S = 6000 m

**2.8 A cricket ball is hit vertically upwards and returns to the ground 6 s later. Calculate**

**(i) Maximum height reached by the ball**

**(ii) initial velocity of the ball (45m, 30 ms-1)**

**Solution:**

Acceleration due to gravity = g = -10 ms^{-1} (for upward motion)

Time to reach maximum height (one-sided time) = t = 6/2 = 3 s

Velocity at maximum height = V_{f} = 0 ms^{-1}

The maximum height reached by the ball S = h =?

The maximum initial velocity of the ball = V_{i }=?

Since,

V_{f }= V_{i} + gxt

V_{i }= V_{f }– gxt

V_{i} = 0 – (-10) x 3

V_{i} = 30 ms^{-1}

Now using 3^{rd} equation of motion

2aS = V_{f2} – V_{i2}

S = ^{ }(V_{f2} – V_{i2})/2a

S = [(0)2 – (30)2/2] x (-10)

S = -90/-20

S = 45 m

**2.9 When brakes are applied, the speed of a train decreases from 96 kmh-1 to 48 kmh-1 in 800 m. How much further will the train move before coming to rest? (Assuming the retardation to be constant). (266.66 m)**

**Solution:**

Initial velocity = V_{i = }96 kmh^{-1} = 96 x 1000/3600 = 96000/3600 = 26.66ms^{-1}

Final velocity = Vf = 48 kmh^{-1} = 48 x 1000/3600 = 48000/3600 = 13.33ms^{-1}

Distance = S = 800 m

Further Distance = S_{1 }=?

First of all, we will find the value of acceleration a

2aS = V_{f}^{2} – V_{i}^{2}

2 x a x 800 = (13.33)^{2} – (26.66)^{2}

1600a = (13.33)^{2} – (26.66)^{2}

1600a = 177.8 – 710.8

1600a = -533

a = -533/1600 = -0.333ms^{-2}

Now, we will find the value of further distance S_{1}:

V_{f }= 0 , S_{1 }=?

2aS = V_{f}^{2} – V_{i}^{2}

2 (-0.333) x S_{1} = (0)^{2} – (13.33)^{2 }

-0.666 S_{1} = -(13.33)^{2}

S_{1} = -177.7/-0.666

S_{1} = 266.66m

**2.10 In the above problem, find the time taken by the train to stop after the application of brakes. (80 s)**

**Solution:**

Initial velocity = v_{i} = 96 kmh^{-1} = 96 x 1000/3600 = 96000/3600 = 26.66ms^{-1}

Final velocity = v_{f} = 0 ms^{-1}

a = -0.333 ms^{-2}

time = t =?

Or v_{f} = v_{i} + at

Or at = v_{f} – v_{i}

t = (v_{f} – v_{i})/a

t = (0 – 13.33) / (-0.333)

t = 2 x 40

t = 80 s

**Class 9 Physics Full Book Numerical Problems with Solution**