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? Solution :
(a) a stone of the same size will have more inertia than a rubber ball.
(b) A train will have more inertia than a bicycle.
(c) A five rupees coin will have more inertia than a one-rupee 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. Solution :
In the given example the velocity of football changes four times. As described below:
(i) when the football player is supplying the force when he kicks the football to another player.
(ii) when the other player kicks football towards the goal.
(iii) When the goalkeeper of other team stops the ball.
(iv) When the goalkeeper kicks the football towards player of his team.
Question 3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch. Solution : Some of the leaves may get detached from a tree if we vigorously shake its branch because the some of the leaves due to property of inertia remain at rest while we vigorously shake branch of the tree as a result those leaves detach and fall off.
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? Solution : when a moving bus brakes to a stop we fall in the forward direction because we are also moving with the speed of bus due to the inertia of motion and when suddenly it puts brakes i.e. comes to rest the lower half of our body also comes to rest but the upper half of our body not being in close contact with bus is still in the phase of motion so we fall in the forward direction.
When the bus accelerates from rest, we are also at rest being on the resting seat as the engine applies force in forward direction we fall backwards due to the inertia now.
Question 5. If action is always equal to the reaction, explain how a horse can pull a cart. Solution : With a balance force the overall impact is absence of movement but with unbalanced forces, the resultant or the bigger force causes the motion. Same is true in the case where a horse pulls a cart. Horse exerts more force on the cart than the cart exerts to resist its movement hence this is an unbalanced force and the cart moves in the direction of horse’s pull.
Question 6. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity. Solution : It is difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity because of the third law of newton when the hose ejects large amounts of water at a high velocity in forward direction the water coming out pushes the hose pipe in backward direction and it becomes difficult to hold it.
Question 7. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s-1. Calculate the initial recoil velocity of the rifle. Solution :
Mass of the rifle,
m1 = 4 kg
Mass of the bullet,
Recoil velocity of the rifle
=v1
Bullet is fired with an initial velocity,
v2 =35 m/s
Initially, the rifle is at rest.
Thus, its initial velocity,
v = 0
Total initial momentum of the rifle and bullet system
= (m1 + m2)v
Total momentum of the rifle and bullet system after firing:
=m1v1 + m2v2 = 4(v1) + 0.05 x 35 = 4v1 + 1.75
According to the law of conservation of momentum:
Total momentum after the firing = Total momentum before the firing
4v1 + 1.75 = 0
v1 = -1.75 / 4 = -0.4375 m/s
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.
Question 8. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s−1 and 1 m s−1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m s−1. Determine the velocity of the second object. Solution :
Mass of one of the objects, m1 = 100 g = 0.1 kg
Mass of the other object, m2 = 200 g = 0.2 kg
Velocity of m1 before collision, v1 = 2 m/s
Velocity of m2 before collision, v2 =1 m/s
Velocity of m1 after collision,v3 = 1.67 m/s
Velocity of m2 after collision =V4
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m1v1 + m2v2 = m1v3 + m2v4
= (0.1)2 + (0.2)1 = (0.1)1.67 + (0.2)v4
= 0.4 = 0.167 + 0.2 v4
so, v4 = 1.165 m/s
Hence, the velocity of the second object becomes 1.165 m/s after the collision.
Question 9. 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 no, provide a reason. Solution : No, it is not possible for the object to be travelling with a non-zero velocity if an object experiences a net zero external unbalanced force since unbalanced forces cannot be equal to zero.
Question 10. When a carpet is beaten with a stick, dust comes out of it. Explain. Solution : When a carpet is beaten with a stick, dust comes out of it because carpet fibres vibrate in forward and backward direction as carpet is beaten but the loosely bound dust particles due to inertia remain at rest and so they come out.
Question 11. Why is it advised to tie any luggage kept on the roof of a bus with a rope? Solution : It is advised to tie any luggage kept on the roof of a bus with a rope because when bus moves the luggage also gets moving with the velocity same as that of the bus and in the same direction but when bus changes direction or deaccelerates, due to inertia of motion luggage moves in the same direction and may get thrown away from roof of buses.
Question 12. 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.
Solution : (c) there is a force on the ball opposing the motion.
Question 13. 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 metric tonnes (Hint: 1 metric tonne = 1000 kg.) Solution :
According to the Question,
initial velocity of truck (u) = 0
distance = s= 400 m and time = 20 s
mass of truck = 7metric tones = 7000kg
400 =0 + 200a
400 = 200a
a = 400/200
therefore, F = m x a = 7000 x 2
= 14000 N
Question 14. A stone of 1 kg is thrown with a velocity of 20m s-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? Solution :
since, v2 = u2 + 2as
0 = 202 + 2 x a x 50(object comes to rest so v=0)
-100a = 400
a = 400/-100 = -4 mm/ s2
therefore, the force of friction between the stone and the ice
= -4 N
Question 15. 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;
(b) the acceleration of the train; and
(c) the force of wagon 1 on wagon 2.
Solution :
(a) The net accelerating force = Force exerted by engine – frictional force of track = 40000 – 5000 = 35000 N
(b) the acceleration of the train = a = F/m = 35000 / (5 x 2000)= 35000/10000 = 3.5 m/s2
(c) the force of wagon 1 on wagon 2
Wagon 1 will have to exert force on all 4 wagons next to it
so mass of other 4 wagons = 2000 x 4= 8000 kg
F = ma = 8000 kg x 3.5 m/s2= 28000 N
Question 16. 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 m S-2? Solution :
since, F=m x a =1500 x -1.7= -2550 N (negative sign symbolises acceleration in opposite direction)
Question 17. 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
Solution : (d) mv
Question 18. 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? Solution : 200 N
Question 19. 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 m s-1 before the collision during which they stick together. What will be the velocity of the combined object after collision? Solution :
Momentum before collision took place
= m1v1 + m2v2
= 1.5 x 2.5 + 1.5 x(-2.5)
= 0
Since the objects stick together after collision hence
momentum after collision
= (m1 + m2) x v
= (1.5 +1.5) x v
= 3v
momentum before collision = momentum after collision
0 = 3v, v= 0/3 = 0
the velocity of the combined object after collision (v)= 0
Question 20. 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. Solution : 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 result is the two opposite and equal forces cancel each other but when one of these forces is bigger than inertia so the object moves in the direction of force applied. As this student explains the truck is massive so the force applied cannot overcome force caused by inertia. Therefore, the truck does not move.
Question 21. A hockey ball of mass 200 g travelling at 10 m s-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 occurred in the motion of the hockey ball by the force applied by the hockey stick. Solution :
mass of hockey ball = 200g = 0.2 kg
v1 = 10 m/s , v2 = -5 m/s (return velocity)
initial momentum of hockey ball = 0.2 kg x 10 m/s= 2 kg m/s
final momentum of hockey ball = 0.2 kg x -5 m/s = -1 kg m/s
change in momentum of hockey ball = 2 – (-1) = 2 + 1 = 3 kg m/s
Question 22. A bullet of mass 10 g travelling 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. Solution :
v =u + at
0 = 150 + a x 0.03 s
a = -150/0.03 = -5000 m/s2
the distance of penetration of the bullet into the block
= 4.5 – 2.25
= 2.25 m
the magnitude of the force exerted by the wooden block on the bullet
m = 10g = 0.01kg
F = m x a = 0.01 kg x -5000 m/s2= -50 N
Question 23. An object of mass 1 kg travelling in a straight line with a velocity of 10 m s-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. Solution :
Wooden block is stationery (at rest) so its velocity (u2) = 0
mass of combined object is = 1 kg + 5 kg = 6 kg
total momentum before the impact = 1 x 10 + 5 x 0= 10 kg m/s
law of conservation of momentum:
total momentum just before the impact = total momentum after the impact= 10 kg m/s
therefore, the velocity of the combined object: 10 = 6 x v = 6v
v = 10/6 = 1.67 m/s
Question 24. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s-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. Solution :
Initial momentum of the object =100 x 5 = 500 kg m/s
Final momentum of the object = 100 x 8 = 800 kg m/s
since v = u + at
8 = 5 + a x 6
a = 3 / 6 = 0.5 m/s2
since F = m x a = 100 x 0.5 = 50 N
Question 25. 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 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. Solution : Since the mass of insect is negligible in comparison to the mass of motorcar therefore there will be no any change in the momentum of motorcar.
Question 26. 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 . Solution :
height from which dumb bell falls
= 80 cm = 0.8 m
since we know
v2 = u2 + 2gh
v2 = 0 + 2 x 10 x 0.8 = 16
v = √16
v = 4 m/s
momentum = mv = 10 x 4 = 40 kg m/s
Question 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example. Solution : Yes, if an object has moved through a distance it can have zero displacement because displacement of an object is the actual change in its position when it moves from one position to the other. So if an object travels from point A to B and then returns back to point A again, the total displacement is zero.
Question 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
Solution :
Distance covered by farmer in 40 seconds
=4 x(10) m= 40 m
Speed of the farmer = distance/time = 40m/40s = 1 m/s.
Total time given in the Question = 2min 20seconds
= 60+60+20 =140 seconds
Since he completes 1round of the field in 40seconds so in he will complete 3rounds in 120seconds (2mins) or 120m distance is covered in 2minutes. In another 20seconds will cover another 20m so total distance covered in 2min20sec
= 120 +20 =140m.
Displacement = √102 + 102 =√200
= 10√ m (as per diagram)
=10 x 1.414= 14.14 m.
Question 3. Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object. Solution : Both (a) as well as (b) are false with respect to concept of displacement.
Question 4. Distinguish between speed and velocity. Solution : Speed of a body is the distance travelled by it per unit time while velocity is displacement per unit time of the body during movement.
Question 5.Under what condition(s) is the magnitude of average velocity of an object equal to its average speed? Solution : If distance travelled by an object is equal to its displacement then the magnitude of average velocity of an object will be equal to its average speed.
Question 6. What does the odometer of an automobile measure? Solution : The odometer of an automobile measures the distance covered by that automobile.
Question 7. What does the path of an object look like when it is in uniform motion? Solution : Graphically the path of an object will be linear i.e. look like a straight line when it is in uniform motion.
Question 8. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is 3 x108ms-1 Solution :
Distance = Speed x time
=
=
Question 9. When will you say a body is in
(i) uniform acceleration?
(ii) non-uniform acceleration? Solution :
(i) uniform acceleration: When an object travels in a straight line and its velocity changes by equal amount in equal intervals of time, it is said to have uniform acceleration.
(ii) non uniform acceleration: It is also called variable acceleration. When the velocity of an object changes by unequal amounts in equal intervals of time, it is said to have non uniform acceleration.
Question 10. A bus decreases its speed from 80kmh-1 to 60km h-1 in 5 s. Find the acceleration of the bus. Solution :
Initial speed of bus (u) =
== =
Question 11. A train starting from a railway station and moving with uniform acceleration attains a speed 40km h-1 in 10 minutes. Find its acceleration. Solution : Since the train starts from rest(railway station) = u = zero
Final velocity of train
=
time (t) = 10 min = 10 x 60
= 600 seconds
Since a = (v – u )/t =
=
Question 12. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object? Solution : If an object has a uniform motion then the nature of distance time graph will be linear i.e. it would a straight line and if it has non uniform motion then the nature of distance time graph is a curved line.
Question 13. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Solution : If the object’s distance time graph is a straight line parallel to the time axis indicates that with increasing time the distance of that object is not increasing hence the object is at rest i.e. not moving.
Question 14. What can you say about the motion of an object if its speed time graph is a straight line parallel to the time axis? Solution : Such a graph indicates that the object is travelling with uniform velocity.
Question 15. What is the quantity which is measured by the area occupied below the velocity-time graph? Solution : The area occupied below the velocity-time graph measures the distance moved by any object.
Question 16. A bus starting from rest moves with a uniform acceleration of 0.1m s-2 for 2 minutes. Find (a)the speed acquired, (b) the distance travelled. Solution :
(a) u=o, , t= 2min = 120 seconds.
v=u+at
so (a) speed acquired
=
(b)
= 720 m.
Question 17. A train is travelling at a speed of 90km h-1. Brakes are applied so as to produce a uniform acceleration of -0.5m s-2. Find how far the train will go before it is brought to rest. Solution :
,
v =0(train is brought to rest)
v= u+at
0 =25 – 0.5 x
0.5t = 25, or t
= 25/0.5
= 50seconds
= 1250 – 625 = 625m
Question 18. A trolley, while going down an inclined plane, has an acceleration of 2cm s-2 . What will be its velocity 3 s after the start? Solution :
, t= 3s
v= u +at = 0 + 2 x 3 = 6 cm/s
Question 19. A racing car has a uniform acceleration of 4cm s-2. What distance will it cover in 10 s after start? Solution :
u = 0, , t= 10 s
= 200 m
Question 20. A stone is thrown in a vertically upward direction with a velocity of 5 cm s-2. If the acceleration of the stone during its motion is 10 cm s-2n the downward direction, what will be the height attained by the stone and how much time will it take to reach there? Solution :
v = 0 (since at maximum height its velocity will be zero)
v = u + at
= 5 + (-10) x t
0 = 5 – 10t
10t = 5, or, t = 5/10 =0.5second.
= 2.5 – 1.25 = 1.25m
Question 21. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s? Solution :
circumference of circular track = 2πr
= 4400/7 m
rounds completed by athlete in
2min20sec = s= 140/40 = 3.5
therefore, total distance covered =4400 / 7 x 3.5= 2200 m
Since one complete round of circular track needs 40s so he will complete 3 rounds in 2mins and in next 20s he can complete half round therefore displacement = diameter = 200m.
Question 22. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C? Solution :
(a) distance = 300m
time = 2min30seconds = 150 seconds
average speed from A to B = average velocity from A to B
= 300m/150s = 2m/s
(b) average speed from A to C = (300+100)m/(150+60)sec
Question 23. Abdul, while driving to school, computes the average speed for his trip to be 20km h-1. On his return trip along the same route, there is less traffic and the average speed is 40km h-1. What is the average speed for Abdul’s trip? Solution :
If we suppose that distance from Abdul’s home to school = x kms
while driving to school :-
,
velocity = displacement/time
20 = x/t, or, t=x/20 hr
on his return trip :-
speed = 40 km h–1 ,
40= x /t’
or, t’ =x/40 hr
total distance travelled = x + x = 2x
total time = t + t’ = x/20 + x/40
=(2x + x)/40 = 3x/40 hr
average speed for Abdul’s trip
= 2x/(3x/40) = 80x/3x = 26.67km/hr
Question 24. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s-2 for 8.0 s. How far does the boat travel during this time? Solution :
since the motorboat starts from rest so u= 0
time (t) = 8s, ,
distance
= 96m
Question 25. A driver of a car travelling at 52 km h-1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h-1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied? Solution :
As given in the figure below AB (in red line) and CD(in red line) are the Speed-time graph for given two cars with initial speeds respectively.
Distance Travelled by first car before coming to rest =Area of
= 325/9 m
= 36.11 m
Distance Travelled by second car before coming to rest =Area of
= 25/6 m
= 4.16 m
∴ Clearly the first car will travel farther (36.11 m) than the first car(4.16 m).
Question 26. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following Questions :
(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?
Solution :
(a) It is clear from graph that B covers more distance in less time. Therefore, B is the fastest.
(b) All of them never come at the same point at the same time.
(c) According to graph; each small division shows about 0.57 km.
A is passing B at point S which is in line with point P (on the distance axis) and shows about 9.14 km
Thus, at this point C travels about
9.14 – (0.57 x 3.75)km
= 9.14 km – 2.1375 km
= 7.0025 km
Thus, when A passes B, C travels about 7 km.
(d) B passes C at point Q at the distance axis which is
=5.28 km
Therefore, B travelled about 5.28 km when passes to C.
Question 27. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10m s-2 , with what velocity will it strike the ground? After what time will it strike the ground? Solution :
Solution :
Let us assume, the final velocity with which ball will strike the ground be ‘v’ and time it takes to strike the ground be ‘t’
Initial Velocity of ball u=0
Distance or height of fall s=20m
Downward acceleration a
As we know,
Or, 2as =
∴ Final velocity of ball, v
t= (v-u)/a
∴ Time taken by the ball to strike= (20-0)/10
= 20/10
= 2 seconds
Question 28. The speed-time graph for a car is shown is Fig. 8.12.
Fig. 8.12
(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Solution :
(a) Distance travelled by car in the 4 second
The area under the slope of the speed – time graph gives the distance travelled by an object.
In the given graph
56 full squares and 12 half squares come under the area slope for the time of 4 second.
Total number of squares = 56 + 12/2 = 62 squares
The total area of the squares will give the distance travelled by the car in 4 second. on the time axis,
Hence the car travels 16.53m in the first 4 seconds.
(b) The straight line part of graph, from point A to point B represents a uniform motion of car.
Question 29. State which of the following situations are possible and give an example for each of these:
(a) an object with a constant acceleration but with zero velocity
(b) an object moving in a certain direction with an acceleration in the perpendicular direction.
Solution :
(a) An object with a constant acceleration can still have the zero velocity. For example an object which is at rest on the surface of earth will have zero velocity but still being acted upon by the gravitational force of earth with an acceleration of 9.81 ms-2 towards the centre of earth. Hence when an object starts falling freely can have constant acceleration but with zero velocity.
(b) When an athlete moves with a velocity of constant magnitude along the circular path, the only change in his velocity is due to the change in the direction of motion. Here, the motion of the athlete moving along a circular path is, therefore, an example of an accelerated motion where acceleration is always perpendicular to direction of motion of an object at a given instance. Hence it is possible when an object moves on a circular path.
Question 30. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth. Solution :
Let us assume an artificial satellite, which is moving in a circular orbit of radius 42250 km covers a distance ‘s’ as it revolve around earth with speed ‘v’ in given time ‘t’ of 24 hours.
= 42250 km
Radius of circular orbit r =42250 x 1000 m
Time taken by artificial satellite t=24 hours
=24 x 60 x 60 s
Distance covered by satellite s =Circumference of circular orbit
,
, t= 10 s
= 4400/7 m
,
Fig. 8.12
= 4/15 m
