9th Class Physics Chapter 2 (Numerical Problems with Solutions)
UNIT No. 2
Kinematics
(Numerical Problems)
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)
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
vf = 20 ms-1 Ans
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
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
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 = ?
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 =?
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).
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
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
Wonderful!
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