9th Class Physics Chapter 2 (Numerical Problems with Solutions)

 

UNIT No. 2

Kinematics

(Numerical Problems)

2.1   A train moves with a uniform velocity of 36 kmh-1 for 10 s. Find the distance travelled by it.                                                                                                  
        Solution                                                                                                                    

Data

velocity of train = v = 36 kmh-1

= 36 km/h

= 36x1000/3600 m/s

= 10 m/s                                                                

                 Time = t = 10 s

Distance travelled by the train = S = ?

Formula

S = vt

Solution

S=vt

Putting values in this equation

S = (10 m/s) (10 s)

         S = 100 m          Ans     
                        

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.?
         Solution
         Data

Initial speed of the train = vi = 0                    (Because initially the  train is at rest)

Distance travelled by the train = S = 1 km = 1000 m

Time = t = 100 s

Speed of train at the end of 100 s = Final speed of the train = vf = ?

Formula

\[S=v_{i}t + \frac{1}{2}at^{2} \]  
\[v_{f}=v_{i} + at\]           
         Solution
         First we wiil calculate the acceleration of the train using 2nd equation of motion  i.e. 
\[S=v_{i}t + \frac{1}{2}at^{2} \] 
         Putting values in this equation
\[1000 =(0))(100) + \frac{1}{2}(a)(100))^{2}\]
\[1000 = 0 +\frac{1}{2}(a)(10000)\]
\[1000 = \frac{1}{2}(a)(10000)\]
\[2000 = (a)(10000)\]
\[\frac{2000}{10000}= a\]
\[a = 0.2 ms^{-2}\] 
        Putting value of acceleration a  in 1st equation of motion i.e. 
\[v_{f}=v_{i} + at\]
\[v_{f}= 0  + (0.2 ms^{-2}))(100 s))\]
\[v_{f}= (0.2 ms^{-2}))(100 s))\]
\[v_{f}= (20 ms^{-1})\]

vf = 20 ms-1 Ans 

        The velocity of train at the end of 100 s is vf = 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.

Solution

Data

Initial velocity of the car= vi = 10 ms-1                 

Acceleration of the car = a = 0.2 ms-2

Time of acceleration = t = half minute = 30 s

Distance travelled by the car in 30 s = S =?

Final velocity of the car = vf =?

Formula

\[v_{f}=v_{i} + at\]

\[S=v_{i}t + \frac{1}{2}at^{2} \]

Solution

First, we will calculate the final velocity of the car using first equation of the motion i.e.

\[v_{f}=v_{i} + at\]

Putting values in this equation

\[v_{f}= (10 ms^{-1} + (0.2 ms^{-2})(30 s)\]
\[v_{f}= (10 ms^{-1} + 6 ms^{-1}\]
\[v_{f}= 16 ms^{-1}\]     
        Final velocity of the car is vf = 16 ms-1    
        Now we will calculate the calculate the distance travelled by the car in 30s using 2nd equation of motion i.e.
\[S=v_{i}t + \frac{1}{2}at^{2} \]
        Putting values in this equation
\[S =(10 ms^{-1})(30 s) + \frac{1}{2}(0.2 ms^{-2})(30))^{2}\]
\[S = 300 m + \frac{1}{2}(0.2 ms^{-2})(900 s_{2})\]
\[S = 300 m + \frac{1}{2}(180 m)\]
\[S = 300 m + 90 m\]
\[S = 390 m\]     
        The distance travelled by the car after 30 s is 390 m.

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 ground?

Solution

Data

Initial velocity of the ball = vi = 30 ms-1                 

Time taken by the ball to reach the highest point = t = 3 s

Final velocity of the ball = the velocity at the highest point = vf = 0

Maximum height reached by the ball = S =?

Time taken by the ball to return the ground = T =?

Acceleration of the ball = a = -g = -10 ms-1

Formula       
\[2aS=v_{f}^{2}-v_{i}^{2}\]       
Solution
First, we will calculate the maximum height reached by the ball using 3rd equation of motion i.e.
\[2aS=v_{f}^{2}-v_{i}^{2}\]   
Putting values in this equation
 \[2aS=v_{f}^{2}-v_{i}^{2}\]
\[2(-10))S= 0^{2} - 30^{2}\] 
\[-20S= 0 - 900\] 
\[-20S=- 900\]
\[S= 45 m\]

Now, we will find the time taken by the ball to reach the ground
Time taken by the ball to return the ground 
\[\left ( Time taken by the ball to retur to groun \right  ) =\left (Time taken by the ball to reach the highest point\right ) + \left (Time taken by the ball from highest point to ground \right )\]    
Time taken by the ball to return to ground = 3 s + 3 s
                              Time taken by the ball to return to ground  = 6 s   Ans 
 

2.5       A car moves with uniform velocity of 40 ms-1 for 5 s. It comes to rest in the next 10 s with uniform deceleration. Find

(i)               deceleration

(ii)             total distance travelled by the car.

Solution

Data

Initial velocity of the car = vi = 40 ms-1          

Time for which car travels at the rate of 40 ms-1 = t1 = 5 s

Time taken by the car for coming to rest from the speed of 40 ms-1 = t2 = 10 s

Deceleration of the car = a =?  

Total distance travelled by the car =?

Formula

Part I

Initial velocity of the car = vi = 40 ms-1

Final velocity of the car = vf = 0

Time taken by the car for coming to rest from the speed of 40 ms-1 = t2 = 10 s

Deceleration of the car = a =?

Formula

\[a= \frac{v_{f}- v_{i}}{t}\]

Putting values in this equation

\[a= \frac{0 - 40}{10}\]

\[a= \frac{- 40}{10}\]

\[a= - 4 ms^{-1}   Ans\]

Part II

We will calculate the distance in three steps.

In 1st step we will calculate the distance S1 for first 5 s when the velocity is unform.

In 2nd step we will calculate the distance S2 for 10 s when the velocity is decreasing.

In 3rd step we will add up these distances S1 and S2 to get the total distance S

Step I

velocity = Initial velocity of the car = vi = 40 ms-1   

tine = Time for which car travels at the rate of 40 ms-1 = t1 = 5 s

Distance travelled by the car in initial 5 s = S1 =?

Formula

S=vt

Solution

S1 = vit1

putting values in this equation

S1 = (40)(5)

S1 = 200 ms-1

Step II

Initial velocity of the car = vi = 40 ms-1   

Final velocity of the car = vf = 0

Deceleration of the car = a = - 4 ms-2

Distance for 10 s when the velocity is decreasing = S2 = ?

Formula

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Solution

\[2aS_{2}=v_{f}^{2}-v_{i}^{2}\] 

Rearrange this equation for S2

\[S_{2}=\frac{(0)^{2}- (40)^{2}}{2(-4)}\] 

\[S_{2}=\frac{0 - 1600}{-8}\] 

\[S_{2}=\frac{- 1600}{-8}\] 

\[S_{2}= 200 m\] 

Total distance travelled by the car = S = S1 + S2

Total distance travelled by the car = S = 200 m + 200 m

Total distance travelled by the car = S = 400 m      Ans 


2.6    A train starts from rest with an acceleration of 0.5 ms-1. Find its speed in kmh-1 when it has moved through 100 m.

Solution

Data

Initial speed of the train = vi = 0 ms-1              (initially the train is at rest) 

Acceleration of the train = a = 0.5 ms-1  

Distance travelled by the train = 100 m

Speed of the train after moving a distance of 100 m = Final speed of the train = vf = ?

         Formula
\[2aS=v_{f}^{2}-v_{i}^{2}\]  
         Solution  
 \[2aS=v_{f}^{2}-v_{i}^{2}\]   
Putting values in this equation
\[2aS=v_{f}^{2}-v_{i}^{2}\] 
\[2(0.5))(100) =v_{f}^{2}-0^{2}\]  
\[100 = v_{f}^{2}- 0 \] 
\[v_{f}^{2} = 100 \] 
\[v_{f} = 10 ms^{-1} \]  
Now we will convert this speed into kmh-1. 
\[v_{f} = \frac{(10)(3600)}{1000} kmh^{-1} \] 
\[v_{f} = 36 kmh^{-1} \] 
Speed of the train after travelling a distance of 100 m is 36 kmh-1
 


2.7      A train staring from rest, accelerates uniformly and attains a velocity 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 the train. 

Solution

Data

Initial speed of the train = vi = 0 ms-1              (initially the train is at rest) 

Time for which train accelerates uniformly = t1 = 2 minutes = 120 s

Speed of train after 2 minutes = 48 kmh-1 = 13.33 ms-1

Time for which train travels with speed of 48 kmh-1 = t2 = 5 minutes = 300 s

Time for uniform retardation of train = t3 = 3 minutes = 180 s  

Total distance travelled by the train = S =?

          Formula         







We will divide this problem into three parts and calculate the distance travelled by the train in each part separately and then add these distances to get the total distance.

Part I

Initial speed of the train = vi = 0 ms-

final speed of the train = vf = 13.33 ms-1

Time = 120 s

Distance travelled by train during this time = S1= ?

Formula

\[v_{f}=v_{i} + at\] 

\[2aS=v_{f}^{2}-v_{i}^{2}\] 

Solution 

First, we will calculate the acceleration of the of the train for this part using 1st equation of motion

\[v_{f}=v_{i} + at\] 

Solving this equation for acceleration a 

\[a=\frac{v_{f}-v_{i}}{t}\]

Putting values in this equation

\[a=\frac{13.33 - 0}{120}\]

\[a=\frac{13.33}{120}\]

\[a= 0.111 ms^{-1}\] 

Now, we will calculate the distance S1 travelled by the train using 3rd equation of motion

\[2aS=v_{f}^{2}-v_{i}^{2}\] 

Solving this equation for S

\[2aS_{1}=v_{f}^{2}-v_{i}^{2}\] 

Putting values in this equation

\[S_{1}=\frac{v_{f}^{2}-v_{i}^{2}}{2a}\] 

\[S_{1}==\frac{(13.33)^{2}-(0))^{2}}{2(0.111)}\]

\[S_{1}==\frac{177.6889- 0}{0.222)}\] 

\[S_{1}==\frac{177.6889}{0.222)}\]

\[S_{1}== 800.4 m\]

Part II

velocity of the train = v = 13.33 ms-1

Time of motion = t = 300 s

Distance travelled by the train during this time = S =?

Formula

S=vt

Solution

S2 =vt

S2 = (13.33) (300)

S2 = (13.33) (300)

S2 = 3999 m

Part III

Initial speed of the train = vi = 13.33 ms-1

final speed of the train = vf = 0 ms-1

Time = 180 s

Distance travelled by train during this time = S2=?

Formula 

\[v_{f}=v_{i} + at\] 

\[2aS=v_{f}^{2}-v_{i}^{2}\] 

Solution

First, we will calculate the acceleration of the of the train for this part using 1st equation of motion

\[v_{f}=v_{i} + at\]

Solving this equation for acceleration a 

\[a=\frac{v_{f}-v_{i}}{t}\]

Putting values in this equation

\[a=\frac{0 - 13,33}{180}\]

\[a=\frac{13.33}{180}\]

\[a= - 0.074 ms^{-1}\]

Now, we will calculate the distance S3 travelled by the train using 3rd equation of motion

\[2aS_{3}=v_{f}^{2}-v_{i}^{2}\] 

Solving this equation for  S3

\[S_{3}=\frac{v_{f}^{2}-v_{i}^{2}}{2a}\] 

Putting values in this

\[S_{3}==\frac{(0)^{2}-(13.33))^{2}}{2(-0.074)}\]

\[S_{3}==\frac{0 - 177.6889}{-0.148}\] 

\[S_{3}==\frac{-177.6889}{-0.148)}\]

\[S_{3}== 1200.6 m\]

Total Distance = S = S1 + S2 + S3

Total Distance = S = 800.4 m + 3999 m + 1200.6 m

Total Distance = S = 6000 m


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

(i) maximum height reached by the ball,

(ii) initial velocity of the ball.

Solution

Data

Time after which ball returns to the ground = T =  6 s

Acceleration of the ball for upward motion = a = -g = -10 ms-1

velocity of the ball at the highest point = final velocity of the ball = vf = 0            

 (when the ball reach at the highest point it comes to rest

Maximum height reached by the ball = S =?

Initial velocity of the ball = vi  = ?

Formula

\[v_{f}=v_{i} + at\]

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Solution

First, we will calculate the initial velocity of the ball using 1st equation of motion

\[v_{f}=v_{i} + at\]

We will consider only the upward motion

As 6s is the time of the whole round trip (i.e., time for both upward and downward motion),

therefore the time for upward motion is half of this time i.e. 3s

Therefore, here we will use t= 3s

Putting values in 1st equation of motion

\[0 =v_{i} + (-10))(3)\]

\[0 =v_{i} - 30\]

\[v_{i} = 30 ms^{-1}\]

Now, we will calculate the maximum height reached by the ball

For this purpose, we will use 3rd equation of motion

\[2aS=v_{f}^{2}-v_{i}^{2}\] 

We will consider the upward motion only

Putting values in 3rd equation of motion

\[2(-10)S= (0)^{2}-(30)^{2}\] 

\[-20S= 0 - 900 \] 

\[-20S= - 900 \] 

\[S= 45 m \] 

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).

Solution

Data

Initial Speed of the train = vi = 96 kmh-1 = 26.67 ms-1

Final Speed of the train = vf = 48 kmh-1 = 13.33 ms-1

         Distance travelled by the train during decrease of speed from 96 kmh-1 to 48 kmh-1 = S1 = 800 m

         How much further will the train move before coming to rest = S2 =?

Formula

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Solution

We will solve this problem in two parts, in first part we will calculate the retardation of the train 

using 3rd equation of motion and in second part we will calculate the distance travelled by the

train before coming to rest

Part I

Now, 1st we will calculate the deceleration of the train using 3rd equation of motion

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Putting values in this equation 

\[2a(800)= (13.33)^{2}-(26.67)^{2}\]

\[(1600)a= (13.33)^{2}-(26.67)^{2}\] 

\[(1600)a= 177.6889 - 711.2889 \]

\[(1600)a= -533.6 \]

\[a= -0.3335 ms^{-2} \]

Now we will calculate the distance travelled by the train using 3rd equation of motion 

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Putting values in this equation

\[2(-0.3335)S= (0)^{2}-(26.67)^{2}\]

\[(-0.667)S= 0 - 711.2889\]

\[(-0.667)S= 711.2889\]

\[S= 1066.4 m \]

Now

Total distance travelled by the train = distance travelled by the train during decrease of speed from 96 kmh-1 to 96 kmh-1  + Further distance travelled by the train before coming to rest 
\[S= S_{1} + S_{2}\] 
\[S_{2} = S - S_{1} \] 
\[S_{2} = 1066.4 - 800 \]
\[S_{2} = 266.4 m \] 
\[S_{2} = 266 m \] 
  

Solution

Data

Initial Speed of the train = vi = 96 kmh-1 = 26.67 ms-1

Final Speed of the train = vf = 48 kmh-1 = 13.33 ms-1

         Distance travelled by the train during decrease of speed from 96 kmh-1 to 48 kmh-1 = S1 = 800 m

          time taken by the train to stop after the application of brakes = t =?

Formula

\[v_{f}=v_{i} + at\]

\[2aS=v_{f}^{2}-v_{i}^{2}\] 

Solution

We will solve this problem in two parts, in first part we will calculate the retardation of the train 

using 3rd equation of motion and in second part we will calculate the time taken by the train to 

stop after the application of brakes.

Part I

\[2aS=v_{f}^{2}-v_{i}^{2}\]

Putting values in this equation 

\[2a(800)= (13.33)^{2}-(26.67)^{2}\]

\[(1600)a= (13.33)^{2}-(26.67)^{2}\] 

\[(1600)a= 177.6889 - 711.2889 \]

\[(1600)a= -533.6 \]

\[a= -0.3335 ms^{-2} \]

Part II

Now, we will calculate the time taken by the train to stop after the application of brakes.

When the train stops its velocity becomes zero. Therefore, our new data becomes

Initial Speed of the train = vi = 96 kmh-1 = 26.67 ms-1

Final Speed of the train = vf = 0

time taken by the train to stop after the application of brakes = t =?

We will calculate the time taken by the train to stop after the application of brakes using 1st 

equation of motion.

\[v_{f}=v_{i} + at\]

We will re arrange this equation to get the value of  

\[t= \frac{v^{f}-v^{i}}{a}\]

Putting values in this equation

\[t= \frac{0 - 26.67}{-0.3335}\]

\[t= \frac{- 26.67}{-0.3335}\]

\[t= 79.97 s\]

By rounding off  the answer 

t = 80 s   Ans













Comments

Post a Comment

Popular posts from this blog

List of Physics Practicals & Model Paper for Physics Practical of SSC

Image
PHYSICS PRACTICALS FOR SSC   Physics Practical Examinations of students of Federal Board of Intermediate and Secondary Education (FBISE), Islamabad are held after the completion of theory Examination of Secondary School Certificate Part-II (SSC-II). This is the combined exam for the Practicals of SSC-I and SSC-II. This exam has two sections namely Section-A and Section-B. Section A This section contains Practicals (experiments) related to the theory topics that students studied in Secondary School Certificate Part-I (SSC-I) or (9 th class). Section B This section contains Practicals (Experiments) related to the theory topics that students studied in Secondary School Certificate Part-II (SSC-II) or (10 th class). LIST OF PRACTICALS Section A 1. To measure the volume of a solid cylinder by measuring length and diameter of a solid cylinder     with Vernier calipers. 2. To measure the thickness of a metal strip or a wire by using a screw gauge. 3....
Read more

List of Physics Practicals & Model Paper for Physics Practical of HSSC

Image
  PHYSICS PRACTICALS FOR HSSC   Physics Practical Examinations of students of Federal Board of Intermediate and Secondary Education (FBISE), Islamabad are held after the completion of theory Examination of Higher Secondary School Certificate Part-II (HSSC-II). This is the combined exam for the Practicals of HSSC-I and HSSC-II. LIST OF PRACTICALS   1. Measure length and diameter of a solid cylinder and hence estimate its volume     quoting proper number of significant figures using Vernier callipers. 2. Measure the diameters of a few ball bearings of different sizes using Screw Gauge     and estimate their volumes. Mention the uncertainty in each result. 3. Determine the weight of a body by vector addition of forces. 4. Verify the two conditions of equilibrium using a suspended meter rod. 5. Determine the moment of inertia of a fly wheel. 6. Verify that the time period of the simple pendulum is directly proportional to the     square roo...
Read more