KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (2022)

Students can Download Science Chapter 9 Force and Laws of Motion Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Science helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka State Syllabus Class 9 Science Chapter 9 Force and Laws of Motion

KSEEB Class 9 Science Force and Laws of Motion Intext Questions and Answers

Question 1.
Which of the following has more inertia?
(a) A rubber ball and a stone of the same size.
Answer :
Stone of the same size

KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (1)

(b) A bicycle and a train.
Answer :
Train

(c) A five rupee coin and a one-rupee coin
Answer :
Five rupee coins.

Question 2.
In the following example, try to identify the number of times the velocity of the ball changes.

A foot ball player kicks a football to another player Of his team who kicks the football towards the goal. The goal keeper of the opposite team collects the football and kicks it towards a player of his own tea. Also, identify the agent supplying the force in each case.
Answer :
Five times
Contact and non-contact forces.

The normal force formula is one type of ground reaction force.

Question 3.
Explain. Why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer :
Inertia, the force tends to influence on its branches by means of non-contact forces the force makes an object to lose its contact force.

Question 4.
Why do you fall in the forward When a moving bus breaks to shop and backward when it acceleration from rest.
Answer :
Inertia adds momentum

Question 5.
If action is always equal to the reaction. Explain how a horse can pull the cart.
Answer :
According to newton’s third law of motion; a forward movement of the muscular energy of Bullocks drags the cart. (Muscles executes contraction and, relaxation)

Question 6.
Explain, why is it difficult for a fireman to hold a hose which ejects large amounts of water at high velocity.
Answer :
An imbalance between kinetic energy of the water force with the momentum of a fireman at its holding point.

Question 7.
From a rifle of mass 4 kg, a bullet of mass 50 g is fired units initial velocity of 35 ms-1 calculate the initial recoil of the velocity of the rifle.
Answer :
m1v1 = m2v2
4 x v1 = 0.05 x 35
= 0.4375 m/s

KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (2)

Question 8.
Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms-1 and 1 ms-1 respectively. They collide and after collection, the first object moves at velocity 1.67 ms-1 determines the velocity of the second object.
Answer :
m1 = 100 g V1 = 2 ms-1
m2 = 200 g V2 = 1 ms-1
Total momentum before the collision
m1u1 + m2u2 = m1v1 + m2v2
0.2 + 0.2 = 0.167 + 0.2
0.4 – 0.167 = 0.25 v2
v2 = 1.165 m/s

KSEEB Class 9 Science Force and Laws of Motion Textbook Exercise Questions and Answers

Question 1.
An object experiences a net zero external unbalanced force. Is it possible for the object to be traveling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Answer :
Greater magnitude force makes the body to move in the direction of the resultant force.

Question 2.
When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer :
Inertia, the force of beaten disturbs the dust particles to move away from the carpet.

Question 3.
Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer :
To retain its position (inertia) the position gets disturbed due to the Variation of the velocity of the body and direction of momentum.

Question 4.
A batsman hits a cricket ball which then rolls on level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a free on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.
Answer :
(c) There is a force on the ball opposing the motion.

Question 5.
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)
Answer :
t = 20 s, V = 400 m a
\(a=\frac{v}{t}=\frac{400}{20}, 20 m s^{-2}\)
F = m
= 7 x 1000 x 20 – 7000 x 20
= 140000 Newtons

KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (3)

Question 6.
A stone of 1 kg is thrown with a velocity of 20 m s~’ across the frozen surface of a lake and comes to rest after traveling a distance of 50 m. What is the force of friction between the stone and the ice?
Answer :
V2 = u + 2as
\(a=\frac{400}{100}=-4 \mathrm{m} / \mathrm{s}^{2}\)
F = ma = 1 x -4 m/s = -4N

Question 7.
An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force;
(b) the acceleration of the train; and
(c) the force of wagon 1 on wagon 2.
Answer :
F = 40000N
f = 5000 N
m = 8000 wagon weigh
= 5 x 2000 = 10000

Question 8.
An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms-2?
Answer :
a – 1.7, ms-2 ; m = 1500 kg; F = ma
KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (4)
= 150 x 17 = 2550 Newton

Question 9.
What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2
(b) mv2
(c) \(\frac{1}{2}\)mv2
(d) mv
Answer :
(d) mv

Question 10.
Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer :
200 kg in/s-2

Question 11.
Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms-1 before the collision during which the stick together. What will be the velocity of the combined object after the collision?
Answer :
Zero

Question 12.
According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Answer :
Force and inertia are imbalanced.

KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (5)

Question 13.
A hockey ball of mass 200 g traveling at 10 ms-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s-1. Calculate the change of momentum that occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer :
Initial momentum = 0.2 x 10 = 2 kg m/s
Find momentum = 0.2 x -5 =-1 kg m/s
Change in momentum = -1- 2 = -3 kg m/s

Question 14.
A bullet of mass 10 g traveling horizontally with a velocity of 150 m s’1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.
Answer :
KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (6)

Question 15.
An object of mass 1 kg traveling in a straight line with a velocity of 10 ms-1 collides with and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Answer :
P = mv = 1 x 10 = 10 kg m/s
m1v1 = m2v2
v2 = 1.67 m/s

Question 16.
An object of mass 100 kg is accelerated uniformly from a velocity of 5
ms-1 to 8 m s-1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Answer :
P1 = mv = 100 x 5 = 50 kg m/s
Find momentum
P2 = m > v
= 100 x 8 = 800 kg m/s
\(F=m a=100\left(\frac{v-u}{t}\right)=100 \times \frac{3}{6}=50 \mathrm{N}\)

KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (7)

Question 17.
Akhtar, Kiran, and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with larger velocity, it exerted, a larger force on the insect. And as a result, the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Answer :
Akhtar’s statement is correct.
m1u1 + m2u2 = m1v1 + m2v2

Question 18.
How much momentum will a dumb¬bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s-2
Answer :
v2 – u2 = 2as
v2 – 02 = 2 x 10 x 0.8
v2 = 16
∴ v = 4 m/s
P = mv = 10 x 4 = 40 m/s

KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (8)

Question 19.
A constant force acts on an object of mass 5 kg for a duration of 2 s. It increases the object’s velocity from 3 m s-1 to 7 m s-1 Find the magnitude of the applied force. Now, if the force was applied for a duration of 5 s, what would be the final velocity of the object?
Answer :
We have been given that u = 3 ms-1 and v = 7 ms-1, t = 2 s and m = 5kg.
From Eq. we have,
\(\mathrm{F}=\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}\)

Substitution of values in this relation gives
\(\mathrm{F}=\frac{5 \mathrm{kg}\left(7 \mathrm{ms}^{-1}-3 \mathrm{ms}^{-1}\right)}{2 \mathrm{s}}=10 \mathrm{N}\)

Now, if this force is applied for a duration of 5 s (t = 5 s), then the final velocity can be calculated by rewriting Eq. as
\(v=u+\frac{F t}{m}\)

On substituting the values of u, F, ni and t, we get the final velocity,
v = 13 ms-1

Question 20.
A motorcar is moving with a velocity of 108 km/h and it takes 4 s. to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg.
Answer :
The initial velocity of the motorcar
u = 108 km/h
= 108 x 1000 m/(60 x 60 s)
= 30 m s-1
and the final velocity of the motorcar
v = 0 m s-1

The total mass of the motorcar along with its passengers = 1000 kg and the time taken to stop the motorcar, t = 4 s. From Eq. we have the magnitude of the force (F) applied by the brakes as
\(\frac{m(v-u)}{t}\)

On substituting the values, we get
F = 1000 kg x (0 – 30) m s-1/4 s
= – 7500 kg m s2 or – 7500 N.

KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (9)

The negative sign tells us that the force exerted by the brakes is opposite to the direction of motion of the motorcar.

Question 21.
A force of 5 N gives a mass m(, an acceleration of 10 m s-2 and a mass m2, an acceleration of 20 ms-2. What acceleration would it give if both the masses were tied together?
Answer :
From Eq. we have
m, = F/a1: and m2 = F/a2

Here,
a1 = 10 ms-2; a2 = 20 m-2 and
F = 5 N

Thus,
m1 = 5N/10 ms-2 = 0.50 kg: and
m1; = 5 N/20 ms-2 = 0.25 kg.

If the two masses were tied together, the total mass, m would be
m = 0.50 kg + 0.25 kg = 0.75 kg.

The acceleration, produced in the combined mass by the 5 N force would be. a = F/m = 5 N/0.75kg = 6.67 ms-2.

Question 22.
The velocity-time graph of a ball of mass 20 g moving along a straight line on a long table is given in Fig.
KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (10)
How much fore does the table exert on the ball to bring it to rest?
Answer :
The initial velocity of the ball is 20 cm s-1. Due to the friction force exerted by the table, the velocity of the ball decreases down to zero in 10 s. Thus, u = 20 cm s-1 ; v – 0 cm s-1 and t =10 s. Since the velocity-time graph is a straight line, it is clear that the ball moves with a constant acceleration. The acceleration a is.
\(\begin{aligned}
a &=\frac{v-u}{t}=\frac{\left(0 \mathrm{cm} \mathrm{s}^{-1}-20 \mathrm{cm} \mathrm{s}^{-1}\right)}{10} \mathrm{s} \\
&=-2 \mathrm{cm} \mathrm{s}^{-2}=-0 \cdot 02 \mathrm{ms}^{-2}
\end{aligned}\)

Question 23.
A bullet of mass 20 g is horizontally fired with a velocity of 150 ms from a pistol of mass 2 kg. Vliet is the recoil velocity of the pistol?
Answer :
W have the mass of the bullet
m1 = 20 g (= 0.02 kg)
and the mass of the pistol.
m2 = 2 kg ;
initial velocities of the bullet (u1) and pistol (u2) 0, respectively the final velocity of the bullet.
V1 = + 150 ni s-1

The direction of the bullet is taken from left to right (positive, by convention. Fig.). Let v be the recoil velocity of the pistol.

Total momenta of the pistol and bullet before the lire, when the gun is at rest
= (2 + 0.02)kg x 0 ins-1
= 0 kg ms-1

Total momenta of the pistol and bullet after it is tired
= 0.02 kg x (+ 150 ms-1) + 2 kg x v m s-1
= (3 + 2v)kg ms-1

According to the law of conservation of momentum. Total momenta after the fire
= Total momenta before the fire 3 + 2v = 0
v = – 1.5 m s-1

The negative sign indicates that the direction in which the pistol would is opposite to that of the bullet, that is, right to left.
KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (11)

KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (12)

Question 24.
A girl of mass 40 kg jumps with a horizontal velocity of 5 in ms-1 onto a stationary cart with frictionless wheels. The mass of the cart is 3 kg. What is her velocity as the cart starts moving? Assume that there is no external imbalanced force working in the horizontal direction.
Answer :
Let v be the velocity of the girl on thecart as the cart starts moving.
The total momenta of the girl and cart before the interaction
= 40 kg x 5rns-1 + 3 kg x 0 ms-1
= 200 kg

Total momènta alter the interaction
= (40 + 3) kg x vms-1
= 43 v kg ms-1

According to the law of conservation of momentum. the total momentum is conserved during the interaction. That is
43 v = 200
\(v=\frac{200}{43}=+4 \cdot 65 \mathrm{ms}\)
The girl on cart would move with a velocity of 4.65 ms-1 in the direction in which the girl jumped (Fig.).
KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (13)
Fig. The girl jumps on to the cart.

Question 25.
Two hockey players of opposite teams, while trying to hit a hockey ball on the ground collide and immediately become entangled. One has a mass of 60 kg and was moving with a velocity 5.0 m s-1 while the other has a mass of 55 kg and was moving faster with a velocity 6.0 m s-1 towards the first player. In which direction and with what velocity will they move after they become entangled? Assume that the frictional force acting between the feet of the two players and the ground is negligible.
Answer :
Let the first player be moving from left to right. By convention left to right is taken as a positive direction and thus right to left is the negative direction. If symbols m and u represent the mass and initial velocity of the two players, respectively. Subscripts 1 and 2 in these physical quantities refer to the two hockey players. Thus,
m1 60 kg ; u1 = + 5 ms-1 ; and
m2 = 55 kg ; u2 = – 6 ms-1.

KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (14)

The total momentum of the two players before the collision
= 60 kg x (+ 5 ms-1) + 55 kg x (- 6 m s-1)
= – 30 kg ms-1

If v is the velocity of the two entangled players after the collision, the total momentum then
= (m1 + m2 ) x v
= (60 + 55) kg x v ms-1
– 115 x v kg ms-1

KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (15)
KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (16)
Equating the momenta of the system before and after the collision, in accordance with the law of conservation of momentum, we get
\(v=-\frac{30}{115}=-0.26 \mathrm{ms}^{-1}\)

Thus, the two entangled players would move with a velocity of 0.26 ms-1 from right to left, which is in the direction the second player was moving before the collision.

KSEEB Class 9 Science Force and Laws of Motion Additional Questions and Answers

Question 1.
It is easier to stop a tennis ball then a cricket ball moving at the same speed. Why?
Answer :
The tennis ball is lighter than the cricket ball, it has smaller momentum hence small force is required to stop.

KSEEB Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (17)

Question 2.
Mention the effects of force.
Answer :
(i) Move a stationary body
(ii) Stop a moving body
(iii) Change the direction of motion of the body
(iv) Change the shape and size of the body.

Question 3.
Mention the three types of inertia.
Answer :
(i) Inertia of rest
(ii) Inertia of motion
(iii) Inertia of direction

Question 4.
State the law of conservation of linear momentum.
Answer :
Total linear momentum after collision is equal to total momentum before collision provided no external unbalanced forces act,

Question 5.
Name any two applications of the law of conservation of linear momentum.
Answer :
(i) Flight of jet planes and rockets
(ii) Recoiling of a gun firing.

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