AP 9th Class Physics 8th Lesson Questions and Answers Force and Laws of Motion

AP State Board new syllabus AP Board Solutions Class 9 Physics 8th Lesson Force and Laws of Motion Questions and Answers.

AP 9th Class Physical Science 8th Lesson Questions and Answers Force and Laws of Motion

9th Class Physics 8th Lesson Force and Laws of Motion Questions and Answers (Exercise)

Question 1.
An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If not, provide a reason.
Answer:
Yes. Object can travel with non-zero velocity without experiencing net zero external unbalanced force in the space. However without unbalanced force the object can’t travel with non – zero velocity on the earth.

Question 2.
When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer:
When a carpet is beaten with a stick it suddenly comes into motion.

• But the dust in it continue to remain in rest due to inertia of rest.
• Hence the dust comes out.

Question 3.
Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer:

i) Backward – due to inertia of rest
ii) Forward – due to inertia of motion
iii) Sideways – due to inertia of direction
That’s why luggage should be tied.

Question 4.
A batsman hits a cricket ball which then rolls on a 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 force 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. Find its acceleration. Find the force acting on it if its mass is 7 tonnes
(Hint : 1 tonne = 1000 kg)
Answer:
Truck: Initial velocity u=0; Displacement s=400m
Time t=20sec ; Mass m=7 tonnes =7000 kg [∵ 1 tonne =1000kg]

Question 6.
A stone of 1 kg is thrown with a velocity of 20 ms^-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Answer:
Mass m =1 kg
Initial velocity u=20ms^-1
Final velocity
v=0 (comes to rest); Displacement
s=50 m
From Kinematics,

Note : Friction force is always opposite to the motion.

Question 7.
A 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 and
b) the acceleration of the train.
Answer:
Mass of enging = 8000 kg;
Mass of each wagon =2000 kg
Force generated by Engine =40000 N; Force of Friction =5000 N
a) Net Accelerating Force =40000-5000=35000 N
b) Acceleration of the train F = ma (Newton’s 2nd law)

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:
Automobile vehicle Visualisation :

Mass m =1500kg
Acceleration a =-1.7ms^-2
F = ma × [By Newton’s 2nd law]
F=1500 × -1.7=-2550N
So, an opposite Force -2550 N should be applied to bring the vehicle to rest.

Question 9.
What is the momentum of an object of mass m, moving with a velocity v ?
a) (mv)²
b) mv²
c) 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:
Cabinet Visualisation:

To move the block with constant velocity across a floor the friction force exerted on the cabinet is -200 N.

Question 11.
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:
The boys reason is not correct. Heavier trucks have greater inertia, so it requires a greater unbalanced force to change its state of rest.

Question 12.
A hockey ball of mass 200 g travelling at 10 ms^-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms^-1. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer:

Question 13.
A bullet of mass 10 g travelling horizontally with a velocity of 150 ms^-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:

Question 14.
An object of mass 1 kg travelling 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:
Block – A
Mass m₁= 1 kg
Initial velocity u₁= 10 ms^-1

Block – B
Mass m₂ = 5 kg
Initial velocity u₂= 0 ms^-1

i) Block A is motion & Block B is at rest.
ii) Block A travelled in straight line & stuck with Block B.
iii) As they are moving together they will have a common velocity.
iv) Total momentum before collision = m₁u₁ + m₂u₂
= 1×10+5×0=10 kgms^-1 or (N)

b) Momentum after Impact = 10 kg ms^-1 (or) (N)
Reason : Momentum before collision is equal to momentum after collision.

c) Combined – velocity: m₁u₁ + m₂u₂ = m₂v+ m₂v
⇒ m₁u₁ + m₂u₂= v(m₁+ m₂)
⇒ 1 x 10 + 5 x 0 = v(1+ 5)

Question 15.
An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms^-1 to 8 ms^-1 in 6s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Answer:
Object:

Mass m = 100 kg Initial velocity u = 5 ms^-1
Final velocity v = 8 ms^-1,Time t, = 6 sec
Initial momentum P₁ = mu = 100 x 5 = 500 kg ms^-1 or (N)
Final momentum P₂ = mv= 100 x 8 = 800 kg ms^-1 (or) (N)

Question 16.
Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expresssway 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 to 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 a 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:
Kiran Akhtar’s reasons are not correct.

• Rahul’s suggestion is correct. Force exerted by windshield and insect are equal opposite according to Newton’s 3rd law.
• But the mass of the insect is lesser than the mass windshield the insect experiences a greater force due to this it dies.

Question 17.
How much momentum will a dumb – bell of mass 10 kg transfer to the floor if it falls from a height of 80cm ? Take its downward acceleration to be 10 ms^-2.
Answer:
Dumb-Bell :

Additional Exercise

Question 1.
The following is the distance-time table of an object in motion

Time in seconds	Distance in metres
0	o
1	1
2	8
3	27
4	64
5	125
6	216
7	343

a) What conclusion can you draw about the acceleration ? Is it constant, increasing, decreasing, or zero ?

Time interval	Distance On)	Final velocity	Change in velocity (m/s)	Acceleration (m/s2)
1	1	1	1-0=1	1/1 = 1
2	8	8	8-1 = 7	7/1 = 7
3	27	27	27-8= 19	19/1=19
4	64	64	64 – 27 = 37	37/1= 37
5	125	125	125-64 = 61	61/1 = 61
6	216	216	216-125 = 91	91/1 = 91
7	343	343	343-216= 127	127/1 = 127

We can observe that the acceleration of the object is increasing.
But it is neither decreasing nor zero nor constant.

Note: This Can be concluded because the distances covered by the object in each successive time interval are not equal. If the acceleration were constant, then the distances covered by the object in successive time intervals would have been in­creasing by a constant amount,

b. What do you infer about the forces acting on the object ?
Answer:
Since the object is in motion, there must be some force acting on it. The distance­time table suggests that the force acting on the object is increasing with time. This is because the acceleration is increasing with time and according to Newton’s Sec­ond Law of Motion, the force acting on an object is directly proportional to its acceleration.

2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 ms^-2. With what force does each person push the motorcar ? (Assume that all persons push the motorcar with the same muscular effort.)
Answer:
1^st Situation:

• When two persons manage to push the motorcar of mass 1200 kg at a uniform velocity along a level road, the net force on the car is zero.
• Let F be the force required to overcome the frictional force acting on the car, then the force applied by each person is F/2 …………….(1)

2^nd situation:

• When three persons push the same motorcar, it accelerates 0.2 m/s².
• Here three persons use F/2 force each [from (1) ]
• The net force acting on the motorcar = 3(F/2) – F = F/2
• By Applying F = ma, we get,
  F/2 = 1200 kg x 0.2 m/s²
  F = 240 N

Hence, the force applied by one person = F/2 = 240/2 = 120 N

3. A hammer of mass 500 g, moving at 50 ms^-1, strikes a nail. The hail stops the ham­mer in a very short time of 0.01 s. What is the force of the nail on the hammer ?
Answer:
1. Here, Mass of the hammer (m) = 500 g = 1/2 kg
Initial velocity of the hammer (u) = 50 m/s
Final velocity of the hammer (v) = 0 m/s
2. Impulse of the hammer on the nail = change in momentum
= mu – mv = m(u – v) = 1/2 (50 – 0) = 25 kg.m/s
3. Impulse of the hammer on the nail = Impulse of the nail on the hammer
4. Hence, impulse of the nail on the hammer = 25 kg m/s
5. Force = impulse / time; given time = 0.01s
F = 25/0.01 s = 2500 N
So, the force of the nail on the hammer is 2500 N (in the opposite direction of the force applied by the hammer)

4. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the mag­nitude of the force required.
Answer:
1. Given that, Mass, m = 1200 kg; Initial velocity, u = 90 km/h = 25 m/s
Final velocity, v = 18 km/H = 5 m/s ; Time, t = 4 s
Acceleration = (final velocity – initial velocity) / time = (5-25)/4 = -5m/s²
(The negative sign indicates that the acceleration is in the opposite direction to the initial velocity)

2. Change in momentum = mass(final velocity – initial velocity)
= 1200 x (5 -25) = -24000 kg m/s
(Again, the negative sign indicates a decrease in momentum)

3. Force = mass x acceleration
= (1200 kg) x (- 5 m/s²) = 1200 x (-5) = – 6000 N
(The negative sign indicates that the force is in the opposite direction to the initial velocity)
So, the magnitude of the force required to slow down the motorcar is 6000 N.

9th Class PS 8th Lesson Questions and Answers (InText)

Question 1.
Which of the following has more inertia: (a) a rubber ball and a stone of the same size ? (b) a bicycle and a train ? (c) a five-rupees coin and a one-rupee coin ?
Answer:
a) A stone has more inertia than a rubber ball.
Reason : Inertia α mass i.e greater the mass greater the inertia, so stone has more mass hence it has more inertia.

b) A train has more inertia than a bicycle.
Reason : Mass of train is more than bicycle.

c) A five rupee coin has more inertia than one rupee coin.
Reason: Mass of ₹ 5 coin is more than ₹ 1 coin.

Question 2.
In the following example, try to identify the number of times the velocity of the ball changes:
“A football player kicks a football to another player of his team who kicks the foot­ball towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”.
Also identify the agent supplying the force in each case.
Answer:
In the foot ball game,
i. Player A kicked to player B -1^st time
ii Player B kicked to Goalkeeper – 2^nd time
iii. Goalkeeper stops the ball – 3^rd time
iv. Goalkeeper kicks to player C – 4^th time
So, velocity of ball changes 4 times.
In (i), (ii), (iii) & (iv) we can observe agents applying force.

Question 3.
Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer:
When we vigorously shake the branches of a tree it comes into motion very quickly.

• But the leaves continue to remain in rest due to inertia of rest.
• Due to this they feel jerks and gets detached from a tree.

Question 4.
Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Answer:
i) Why do we fall in forward direction when a moving bus brakes to a stop suddenly:
Reason:

• A person in the moving bus is in the state of motion.
• When the bus stops suddenly.
• The feet of the person which are in contact with the floor of the bus comes to rest suddenly.
• But the upper part of the body continue to remain in motion due to inertia of motion.
• That’s why we fall in the forward direction.

ii) Why fall in backward direction when a bus starts suddenly: do we
Reason:

• When the bus is at rest the person sitting in the bus also at rest.
• When it moves suddenly the feet in contact with the bus comes to motion.
• But the upper part of the body continue to remain in rest due to inertia ot rest.
• That’s why we tend to fall in the backward direction.

Examples:

Question 1.
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 ms^-1 to 7 ms^-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 ms¹ and v = 7 ms^-1  t = 2 s and m = 5 kg.
From Eq. (8.5), we have, \(F=\frac{m(v-u)}{t}\)
Substitution of values in this relation gives F = 5 kg (7 ms^-1 – 3 ms¹ )/2 s = 10N.
Now, if this force is applied for a duration of 5 s (t = 5s), then the final velocity can be calculated by rewriting Eq. (8.5) as v = u + \(\frac{\mathrm{Ft}}{\mathrm{~m}}\)
On substituting the values of u, F, m and t, we get the final velocity, v = 13 ms^-1

Question 2.
Which would require a greater force – accelerating a 2 kg mass at 5 m s^-2 or a 4 kg mass at 2 ms^-2 ?
Answer:
From Eq. (8.4), we have F = ma.
Here we have m₁ = 2 kg; a₁ = 5 ms^-2 and m₂ = 4 kg ; a₂ = 2 ms^-2
Thus, F₁= m₁a₁ = 2 kg x 5 ms^-2 = 10 N and F₂ = m₂a₂ = 4 kg x 2 ms^-2 = 8 N.
⇒ F₁ > F₂
Thus, accelerating a 2 kg mass at 5 ms^-2 would require a greater force.

Question 3.
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 ms^-1 and the final velocity of the motorcar v = 0 ms^-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. (8.5),
we have the magnitude of the force (F) applied by the brakes as m(v – u)/t.
On substituting the values, we get F = 1000 kg x (0 – 30) m s^-1 /4 s
= -7500 kg ms^-2 or -7500 N.
The negative sign tells us that the force exerted by the brakes is opposite to the direction of motion of the motorcar.

Question 4.
A force of 5 N gives a mass m₁ an acceleration of 10 m^-2 and a mass m₂, an acceleration of 20 m s^-2. What acceleration would it give if both the masses were tied together ?
Answer:
From Eq. (8.4) we have m₁ = F/a₁ and m₂ = F/a₂
Here, a₁ = 10 ms^-2; a₂ = 20 ms^-2 and F = 5 N.
Thus, m₁ = 5 N/10 m s^-2 = 0.50 kg and
m₂ = 5 N/20 m s^-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, a produced in the combined mass by the 5 N force would be,
a = F/m = 5 N/0.75 kg = 6.67 m s^-2

Question 5.
The velocity-time graph of a ball of mass 20 g moving along a straight line on a long table is given in the adjacent figure. How much force 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 frictional 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 \(a=\frac{v-u}{t}\)
= (0 cm s^-1 – 20 cm s^-1)/10 s = – 2 cm s^-2 = – 0.02 m s^-2.
The force exerted on the ball F is, F = ma = (20/1000) kg x (- 0.02 m^-2)
= -0.0004 N.

The negative sign implies that the frictional force exerted by the table is opposite to the direction of motion of the ball.

9th Class Physical Science Chapter 8 Questions and Answers (Lab Activities)

(Page No. 8) Activity 8.1

Question 1.
1) Make a pile of similar carom coins on a table, as shown in below figure.

2) Attempt a sharp horizontal hit at the bottom of the pile using another carom coin or the striker. If the hit is strong enough, the bottom coin moves out quickly. Once the lowest coin is removed, the inertia of the other coins makes them ‘fall’ vertically on the table.
Answer:
Activity explains about Inertia.

(Page No. 10) Activity 8.2

Question 2.
1) Set a five – rupee coin on a stiff card covering an empty glass tumbler standing on a table as shown in adjacent figure.
2) Give the card a sharp horizontal flick with a finger.

If we do it fast, then the card shoots away, allowing the coin to fall vertically into the glass tumbler due to its inertia.
3) The inertia of the coin tries to maintain its state of rest even when the card flows off.
Answer:
Activity explains about Inertia.

(Page No. 10) Activity 8.3

Question 3.
1) Place a water – filled tumbler on a tray.
2) Hold the tray and turn around as fast as you can.
3) We observe that the water spills. Why?
Answer:
Water spills due to inertia of rest & motion.

(Page No. 20) Activity 8.4

Question 4.
1) Request two children to stand on two separate carts as shown in adjacent figure.
2) Give them a bag full of sand or some other heavy object. Ask them to play a game of catch with the bag.
3) Does each of them experience an instantaneous force as a result of throwing the sand bag?
4) You can paint a white line on cartwheels to observe the motion of the two carts when the children throw the bag towards each other.
Answer:
1) Yes, the cart experiences an instantaneous force.
2) They move further back according to Newton’s third law.