Maharashtra Board Class 9 Science Solution Chapter 1 – Laws of Motion
Balbharati Maharashtra Board Class 9 Science Solution Chapter 1: Laws of Motion. Marathi or English Medium Students of Class 9 get here Laws of Motion full Exercise Solution.
Std |
Maharashtra Class 9 |
Subject |
Science Solution |
Chapter |
Laws of Motion |
1.) Match the first column with appropriate entries in the second and third columns and remake the table.
1) Negative acceleration : The velocity of the object decreases: A vehicle moving with the velocity of 10 m/s, stops after 5 seconds.
2 Positive acceleration : The velocity of the object increases: A car, initially at rest reaches a velocity of 50 km/hr in 10 seconds
3) Zero acceleration : The velocity of the object remains constant: A vehicle is moving with a velocity of 25 m/s
2.) Clarify the differences
A.) Distance and displacement
Distance:-1) Distanceis the initial and final position path length.
2) Distance is always positive as well as zero but not negative.
3) Distance has only magnitude and not direction and hence this is scalar quantity.
4) Distance is depend on the path followed.
5) Ex. Manisha cover 20 meters from home.
Displacement:-1) It is shortest path length between initial and final position.
2) It is positive as well as negative.
3) It has both magnitude and direction and hence this is vector quantity.
4) It is depends on initial and final position.
5) Ex.if an body moves from one position to another position.
B.) Uniform and non-uniform motion.
Uniform motion:-1)If uniform motion covers same distance in equal interval of time then this body is called as uniform motion.
2) In uniform motion graph between distance and time is straight line.
3) uniform motion has zero acceleration.
4) uniform motion having constant speed.
5) Ex.electron revolve around the nucleus.
Non-uniform motion:- 1) If Non-uniform motion covers unequal distance in equal interval of time then this body bodyis called as Non-uniform motion.
2) In non-uniform motiongraph between distance and time is not straight line.
3) non-uniform motion has non-zero acceleration.
4) non-uniform motionhaving variable speed.
5) Ex.bouncing ball.
3.) Complete the following table.
u (m/s) | a (m/s2) | t (sec) | v = u + at (m/s) |
2 | 4 | 3 | —- |
—- | 5 | 2 | 20 |
Ans- v = u + at (m/s)
v =2+4*3=2+12=14
v=14
v = u + at (m/s)
20=u+5*2
u=20-10
u=10
u (m/s) | a (m/s2) | t (sec) | s = ut +1/2 at2 (m) |
5 | 12 | 3 | —- |
7 | —- | 4 | 92 |
Ans- s = ut +1/2 at^{2}
^{S= 5*3+1/2*12*3*3=15+54}
^{S=69}
s = ut +1/2 at^{2}
^{92=7*4+1/2*a*4*4}
^{92-28=16/2*a}
^{a=64*2/16}
^{a=8}
^{ }u (m/s) | a (m/s^{2}) | S(m) | v ^{2}= u^{2} + 2as (m/s)2 |
4 | 3 | —- | 8 |
—- | 5 | 8.4 | 10 |
Ans- v ^{2}= u^{2} + 2as
8*8=4*4+2*3*s
S= 64-16/6
S=48/6
S=8
v ^{2}= u^{2} + 2as
10*10= u^{2}+2*5*8.4
u^{2=}100-84
u=4
4.) Complete the sentences and explainthem.
a.)The minimum distance between the start and finish points of the motion of an object is called the displacement of the object.
Explanation:-Displacement is shortest path length between initial and final position.
b.) Deceleration is negative acceleration.
Explanation:- In negative accelretion decreases the velocity of an object.
c.) When an object is in uniform circular motion, its velocity changes at every point.
Explanation:-its covers same distance in equal interval of time then this body is called as uniform motion. uniform circular motion having constant speed changing with velocity.
d.) During collision total momentum remains constant.
Explanation:-before collision and after collision are equal means total momentum is conserved.
e.) The working of a rocket depends on Newton’s thired law of motion.
Explanation:-in Newton’s third law of motion each an every action there is equal and opposite reaction.
5.) Give scientific reasons.
a.) When an object falls freely to the ground, its acceleration is uniform.
Ans-When an object falls freely to the ground, its acceleration is uniform because the velocity increases by equal amount in equal interval of time.and the any body fall on the earth surface due to gravitational force of attraction.
b.) Even though the magnitudes of action force and reaction force are equal and their directions are opposite, their effects do not get cancelled.
Ans- Even though the magnitudes of action force and reaction force are equal and their directions are opposite, their effects do not get cancelled.because same object their effects cancel to each other and different object not cancel to each other.
c.) It is easier to stop a tennis ball as compared to a cricket ball, when both are travelling with the same velocity.
Ans- It is easier to stop a tennis ball as compared to a cricket ball, when both are travelling with the same velocity because in this case mass of cricket ball is more than tennis ball and hence momentum of cricket ball is more than the tennis ball.
d.) The velocity of an object at rest is considered to be uniform.
Ans- The velocity of an object at rest is considered to be uniform because,at rest there is zero displacement and hence they do not change with time.
6.) Take 5 examples from your surroundings and give explanation based on Newtons laws of motion.
1) Man driving the car.-Newtons first law of motion.because inertia of rest and inertia of motion.
2) Any body which placed on a plane surface of earth. – Newtons first law of motion.because tendency of an body to mange state of rest.
3) frog is swimming.-Newtons third law of motion.because frog is push the water. back and the water push its body forword.
4) 9kg and 10 kg books put on table.-Newtons second law of motion. Because less mass having more acceleration.
5) cricketer catching a ball.-Newtons second law of motion. Because greater the rate of change of momentum greater the force.
7.) Solve the following examples.
a) An object moves 18 m in the first 3s, 22 m in the next 3 s and 14 m in the last 3 s. What is its average speed? (Ans: 6 m/s)
Ans-d_{1}=18 m , d_{2}=22 m, d_{3}=14 m
t_{1}=3s , t_{2}=3s, t_{3}=3s
average speed=total distance travelled/total time taken
=d_{1}+d_{2}+d_{3}/t_{1}+t_{2}+t_{3}
= 18+22+14/3+3+3
=54/9
=6m/s
b) An object of mass 16 kg is moving with an acceleration of 3 m/s^{2 }.Calculate the applied force. If the same force is applied on an object of mass 24 kg, how much will be the acceleration? (Ans: 48 N, 2 m/s2)
Ans-m_{1}=16 kg, m_{2}=24 kg, a_{1}=3 m/s^{2}, a_{2}=?
F=m_{1}a_{1}=16*3=48N
F=m_{2}a_{2}
A_{2}=F/m_{2}
A_{2}=48/24=2 m/s^{2}
c) A bullet having a mass of 10 g and moving with a speed of 1.5 m/s,penetrates a thick wooden plank of mass 900 g. The plank was initially at rest. The bullet gets embedded in the plank and both move together. Determine their velocity.
(Ans: 0.15 m/s)
Ans-Given data m_{1}=10gm=10*10^{-3}kg,m_{2}=90gm=90*10^{-3} kg
u_{1}=1.5m/s, u_{2}=0m/s, v=?
we know that law of conservation of momentum,
m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}
plank was initially at rest i.e u_{2}=0m/s
m_{1}u_{1}=m_{1}v_{1}+m_{2}v_{2}
but v_{1=} v_{2=} v
v= m_{1}u_{1/} m_{1+} m_{2}
v=10*10^{-3}*1.5/90*10^{-3 +}10*10^{-3 }
v=15/100
v=0.15m/s
d) A person swims 100 m in the first 40 s, 80 m in the next 40 s and 45 m in the last 20 s. What is the averagespeed? (Ans: 2.25 m/s^{2})
Ans-d_{1}=100 m, d_{2}=80 m ,d_{3}=45 m
t_{1}=40 s ,t_{2}=40 s ,t_{3}=20 s
total distance covererd/total time taken=d_{1}+ d_{2}+ d_{3}/t_{1}+t_{2}+t_{3}
=100+80+45/40+40+20
=225/100
2.25 m/s.
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