Q: A driver of a car travelling at 52km h−1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5s. Another driver going at 3km h−1 in another car applies his brakes slowly and stops in 10s. 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?

(A) : Initial speed of the car A u=52 kmh−1 =52×185 = 14.44 ms−1v=0t=5With the help of these initial and final points for both the cases, we can plot the graph of speed vs time.Area under the speed-time graph gives the distance covered.u=3kmh−1 =3×185 =0.83Thus car A travels more distance than B.

Question 1

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?

Accepted Answer

Given :      Diameter of the circular track   D=2R=200  m                  Time     t=2 utes 20 s  =140  sAs the athlete completes one round in 40 seconds, thus he will be at point B after 140  seconds by covering  three and half  rounds.∴  Displacement of the athlete       s′=2R=200 mDistance covered     d=3.5×(2π R)d=3.5×722​×200 =2200   m

Question 2

Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 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?

Accepted Answer

(a) From A to B,Average velocity=Total displacement/Total timevaverage​=150300​=2m/secso velocity will be 2 m/sec in rightward direction.Average speed=Total distance/time            =300/150=2 m/s(b)From A to C,Average velocity=Total displacement/Total time=200/210=0.952 m/sAverage Speed=totaltimetakentotaldistancetravelled​             =210400​=1.905 m/sec

Question 3

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 30km h−1. What is the average speed for Abdul's trip?

Accepted Answer

Let us assume that Abdul is travelling x kilometer distance while going to school.Distance = Speed x Time1)When it's average speed is 20 km/hrx=v1​×t1​=20×t1​2)When it's average speed is 30 km/hrx=v2​×t2​=30×t2​Vaverage​=totaltimetakentotaldistancetravelled​Vaverage​=v1​x​+v2​x​ x+x​=20x​+30x​ 2x​=24 km h−1

Question 4

A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0m/s2 for 8.0s. How far does the boat travel during this time?

Accepted Answer

We know that Distance is given by second equation of motion , S=ut+21​at2Here S=?=Distanceu=0 = initial velocitya=3m/s2 = acceleratint=8s= timeSo, using the equation we getS=ut+21​at2⇒S=0+21​×3×82⇒S=96m

Question 5

A driver of a car travelling at 52km h−1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5s. Another driver going at 3km h−1 in another car applies his brakes slowly and stops in 10s. 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?

Accepted Answer

(A)   :         Initial speed of the car A      u=52  kmh−1 =52×185​ = 14.44  ms−1v=0t=5With the help of these initial and final points for both the cases,  we can plot the graph of speed vs time.Area under the speed-time graph gives the distance covered.u=3kmh−1 =3×185​ =0.83Thus car A travels more distance than B.

Question 6

Fig 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?

Accepted Answer

(a) The slope of distance-time graph gives the speed of the objects.            As the slope of object  (B) is the greatest among all, thus  B has the largest speed and hence it is the fastest among all the objects.(b)  :      As the curve of the objects do not intersect with each other even once, hence they cannot be at the same point on the road at the same time.(c)   :      On the y axis  (i.e  distance line), 7 small represents 4 km       ∴  1 small line  =74​ =0.57 km            Position of C at the instant B passes A is  8 km       (d)   ;     Position of B at the instant it passes C is 2nd small line above 4 km i.e 4+2×0.57 =5.14 km         ∴ Distance traveled by B   =5.14−0 =5.14 km

Question 7

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s−2, with what velocity will it strike the ground? After what time will it strike the ground?

Accepted Answer

Height, s=20mAcceleartion , a=10m/s2By 3rd equation of motion v2=u2+2asv2=0+2×10×20=400⇒ v=20m/sSo, we conclude final velocity , v=20m/sagain, by 1st equation of motion , v=u+at⇒t=av−u​⇒t=1020​=2sSo, it will strike the ground after 2s

Question 8

The speed-time graph for a car is shown in Fig.(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?

Accepted Answer

(a)The total number of divisions on horizontal axis for 2 s is 5.The total number of divisions on vertical axis for 2 m/s is 3.The total displacement of the car during these two seconds will be given as the area under all these squares.The area under the (3×5) squares is given as:A15​=2×2 = 4So, the area under one small square is :A1​=154​Now, the area under the graph for 4 sec is:the number of squares under the graph is :n=57 small squares + 6 half small squaresSo, total number of squares =60.So, the distance traveled by the car is the area under the curve.S=60×154​S=16 m(b)The car moves under the uniform motion after the instant t=6 s because after this instant, the speed of the car becomes constant.

Question 9

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.

Accepted Answer

(a) Let say a stone is thrown in vertical direction with initial speed u. As the time passes its velocity decreases and at the maximum height it attained, its velocity is zero but it has an acceleration (9.8) m/s2 in the downward direction due to gravity. (b) When a pebble is tied with an end of string and it is whirled in horizontal circular motion with the other end of the string at the centre of circular path, then it has a constant magnitude of centripetal acceleration towards the centre. Hence the acceleration is in perpendicular direction to the motion of pebble.

Question 10

An artificial satellite is moving in a circular orbit of radius 42250km. Calculate its speed if it takes 24 hours to revolve around the earth.

Accepted Answer

Time taken by the satellite to revolve around the earth once      t=24 hours ∴  t=24×3600 s =86400  sRadius of the orbit       R=42250  km =42250000 mLet the velocity of the satellite be   v∴    v=t2πR​=864002×3.14×42250000​⟹ v=3070.95  ms−1  =3.1  kms−1

Question 11

An object experience 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.

Accepted Answer

Yes, an object may travel with a non-zero velocity even when the net external force on it is zero. An object moving in some direction with constant velocity will continue in its state of motion as long as there are no external unbalanced forces acting on it. In order to change the motion of the object, some external unbalanced force must act upon it.A raindrop falls down with a constant velocity. The weight of the drop is balanced by the up thrust and the velocity of air. The net force on the drop is zero.

Question 12

When a carpet is beaten with a stick, dust comes out of it. Explain.

Accepted Answer

When a carpet is beaten with a stick, the dust comes out of it because of law of inertia.Initially the dust particles are at rest along with the carpet. Beating the carpet with the stick makes the carpet move but the dust particles remain at rest due to inertia of rest, thus the dust gets detached from the carpet.

Question 13

Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Accepted Answer

When the bus moves in forward direction, the luggage kept on the roof also moves along with the bus in forward direction. Now when the bus suddenly stops, it comes to rest but the luggage is still in motion in forward direction due to inertia of motion, thus the luggage falls off the roof of bus.Hence it is advised to tie it.

Question 14

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 a stop because:

Accepted Answer

When the ball rolls on the ground, the ground exerts a friction force on the ball in the opposite direction of its motion. Due to this friction force, the ball retards (gets slowed down) and finally stops after some time covering small distance.

Question 15

A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400m in 20s. Find the acceleration. Find the force acting on it if its mass is 7 tonnes. (Hint: 1 tonne = 1000 kg.)

Accepted Answer

Given data:Mass of stone(m)=7000 kgDistance(S)=400 meterTime(t)=20 second1) We have to find the acceleration of the truckSo from the equation of motion S=ut+21​a×t2, whereu=0 S=21​a×t2a=t22×S​=20×202×400​=2sec2m​2) Force acting on truckSo,from newton's 2nd law of motionF=ma=7000×2=14000N=14kN

Question 16

A stone of 1kg 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 50m. What is the force of friction between the stone and the ice?

Accepted Answer

Before we proceed we have to make a free body diagram of the stone-ice system. In the figure, we are seeing that frictional force is acting in leftward while the stone in moving rightward direction. One thing we have to fix in our d that frictional force will always act opposite to the direction of motion of the body.Now on resolving the forces,f−ma or f=ma                         ------(1)also form the equation of motion v2=u2+2as, here final velocity is zero,so v=0u2=−2asa=2s−u2​=−2×50202​=−4sec2m​ on substituting the given data in equation(1)f=1×(−4)=−4N

Question 17

A 8000kg engine pulls a train of 5 wagons, each of 2000kg, along a horizontal track. If the engine exerts a force of 40000N and the track offers a friction force of 5000N, then calculate:(a) The net accelerating force(b) The acceleration of the train(c) The force of wagon 1 on wagon 2

Accepted Answer

Given :    Mass of engine    M=8000 kg                Mass of each wagon    m=2000  kg                Frictional force acting in backward direction    f=5000  N(a)  :   The net force acting on the train  F′=F−f =40000−5000=35000 N(b)  :     Let the acceleration of the train be  a         ∴    F′=(5m+M) a            35000= (5×2000+8000) a          ⟹ a=1.944  ms−2(c)   :    External force is applied directly to wagon 1 only. The net force on the last four wagons is equal to the force applied by wagon 2 by wagon 1.             Let the acceleration of the wagons be  a′              35000 =(5m)a′            ⟹ 35000=10000×a′           Acceleration of the wagons      a′=3.5 ms−2           Mass of last 4 wagons     m′=4×2000 kg      ∴  Net force on last 4 wagons     F1​=8000×3.5 =28000 NThus force on wagon 2 by wagon 1 is 28000 N.

Question 18

An automobile vehicle has a mass of 1500kg. 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?

Accepted Answer

Given data:mass of vehicle=1500kgRetardation=1.7m/{sec2Here we can define this negative acceleration as retardation,which state that body is accelerating in opposite direction of applied force.from newton's 2nd law of motionf=mawhere f is the frictional force actin between the vehicle and the road,which will always act opposite to the direction of motion of body.f−ma=0f=ma=1500×1.7=2550N

Question 19

What is the momentum of an object of mass m, moving with a velocity v?

Accepted Answer

Momentum of an object is defined as the product of its mass and the velocity with which the object is moving.Let velocity of the object be  v and its mass be  m∴Momentum, P=mv

Question 20

Using a horizontal force of 200N, 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?

Accepted Answer

For an object to move with constant velocity, net external force experienced by it must be equal to ZERO          i.e        Fexternal​ =0Thus a friction of 200 N must be exerted on the cabinet so that it moves with a constant velocity under the applied horizontal force of 200 N.

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