Question-1
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edges of a table.
Solution:
(a) As the size of railway carriage is very small as compared to the distance between two stations, hence the carriage can be treated as a point object.
(b) The monkey can be considered as point object because its size is very small as compared to distance covered by the cyclist.
(c) Because the size of the spinning cricket ball is comparable with the distance through which the ball may turn on hitting the ground, hence the cricket ball cannot be considered as a point object.
(d) The size of beaker is not negligible as compared to the height of the table. Hence, a beaker slipping off the edges of a table cannot be treated as a point object.
Question-2
Choose the correct entries in the brackets below:
a. (A/B) lives close to the school than (B/A)
b. (A/B) starts from the school earlier than (B/A)
c. (A/B) walks faster than (B/A)
d. A and B reach home at the (same/different) time.
e. (A/B) overtakes (B/A) on the road (once/twice).
Solution:
a. â€˜Aâ€™ lives close to the school than â€˜Bâ€™.
b. A starts from the school earlier than B.
c. B walks faster than A.
d. A and B reach home at the same time.
e. B overtakes A on the road once.
Question-3
Solution:
The time taken by the woman to return home from office
=(2.5 Ã—1000 Ã— 3600)/(25 Ã— 1000) = 360 sec or 6 min
Taking 9 am is chosen as origin for the time axis, and home for the distance axis, the x - t graph for the woman's motion is shown in figure.
Question-4
Solution:
By calculation:
Steps |
Distance moved (m) |
Time (s) |
5 |
5 |
5 |
5 + 8 = 13 |
5 + 2 = 7 |
5 + 8 = 13 |
13 + 8 = 21 |
7 +2 = 9 |
13 + 8 = 21 |
21 + 8 = 29 |
9 + 2 = 11 |
21 + 8 = 29 |
29 + 8 = 37 |
11 + 2 = 13 |
29 + 8 = 37 |
Question-5
Solution:
Take the direction of motion of the plane as positive.
Velocity of plane with reference to earth = Vpe = 500 km/h
Velocity of smoke with reference to earth = Vse = x (to be found out)
Velocity of smoke with reference to plane = Vsp = -1500 km/h
âˆ´Vsp = Vse - Vpe
Now, -1500 = x - (500)
x = -1000 km/h
(-ve sign indicates the direction opposite to that of the plane's motion).
Question-6
Solution:
We know that the equation for velocity at t sec is given by,
v^{2}(t) = v^{2}(0) + 2a [x(t) - x(0)]
where
The displacement [x(t) - x(0)] = 200 m
Velocity at time t = v(t) = 0
Velocity at time t is 0 = v(0) = 126 km/h = 35 m/s
Therefore the equation becomes
0 = 35^{2} + 2a(200)
âˆ´ Acceleration a = - [35^{2} / 400]
= - 3.06 m/s^{2}
(Negative sign indicates that the acceleration is opposite to the direction of motion of the car)
v(t) = v(0) + a t
0 = 35 - 3.06 t
Time for stopping t = 11.4 s.
Question-7
Solution:
[Original distance between the guard of B and the driver of A = Distance travelled by B in 50 s - Distance travelled by A in 50 s. This approach will yield the same result. This method is to be necessarily used if both the trains had been moving with different accelerations]
Distance travelled by B in 50 seconds:
x_{2} = v(0)t + Â½ at^{2}
= 20 Ã— 50 + Â½ Ã— 1 Ã— (50)^{2}
= 2250 m
Distance travelled by A in 50 secs:
x_{1} = v(0)t + 0 = 20 Ã— 50
= 1000 m
Therefore x_{2} - x_{1} = Distance between the guard of B and driver of A
= 2250 - 1000 = 1250 m.
Question-8
Solution:
Since the common point of consideration is A, consider all the velocities relative to A.
Then time for B to reach A with its relative velocity and acceleration should be at least the time taken by C to reach without any acceleration. V_{a }= 54 -36 km/h
= 18 km/h = 5 m/s
Direction of motion of A is taken as positive.
Velocity of B w.r. to A = V_{b }- V_{a} = 54 - 36 km/h.
Velocity of C w.r. to A = V_{c} - V_{a }= 54 - (-36) km/h = 90 km/h = 25 m/s.
Time for C to reach A at 90 km/h relative velocity = 1000/25 m
= 40 s
Considering B, v(0) = 5 m/s; t = 40 s,
The displacement x(40) â€“ x(0) = 1 km = 1000 m
x(0) â€“ x(t) = v(0).t + Â½ a t^{2}
1000 = 5 Ã— 40 + Â½ a Ã— 40 Ã— 40
âˆ´a = 1 m/s^{2} = the required acceleration.
Question-9
Solution:
Let the speed of the busses be V_{b} and that of the cycle V_{c}.
V_{c} =20 km/h = 3 km/min
Velocity of the bus from B relative to the cycle = V_{b} + V_{c}
Velocity of the bus from A relative to the cycle = V_{b} - V_{c}
Distance between two busses coming opposite to the cycle = [(V_{b} + V_{c}) Ã— 6] km
Distance between two busses coming in the same direction = [(V_{b} - V_{c}) Ã— 18] km
ie V_{b }= 2 Ã— V_{c}
= 2 Ã—1/3 km/min
= 2/3 km/min
Speed of the buses = 2/3 km/min = 40 km/h
Distance between two busses = (V_{b} + V_{c}) Ã—6 km
= (2/3+1/3) Ã—6 km = 6 km.
T = time for covering the distance between the two busses
= 6 /(2/3)min.
= 9/2 min.
Question-10
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocities and accelerations of the ball at the highest point of its motion?
(c) Choose x = 0 and t = 0s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis and give the signs of positions, velocity and acceleration of the ball during its upward and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player's hand?
(Take g = 9.8 ms^{-2} and neglect air resistance).
Solution:
a.Acceleration acting on the ball is only the acceleration due to gravity.
Direction remains to be vertically downwards throughout the flight of the ball.
b. The velocity of the ball at the highest point will be zero.
The acceleration still continues to be due to gravity directed downwards and is equal to 9.8 m/s^{2}.
c.
Direction of motion |
Position |
Velocity |
Acceleration due to gravity |
Upward Motion |
Positive |
Negative |
Positive |
Downward Motion |
Positive |
Positive |
Positive |
Question-11
a. with zero speed at an instant may have non-zero acceleration at that instant,
b. with zero speed may have non-zero velocity
c. with constant speed must have zero acceleration
d. with positive value of acceleration must be speeding up.
Solution:
a. Acceleration may increase or decrease the velocity till zero and continue to act on it. Hence zero velocity is possible with non-zero acceleration. Example is a ball thrown up in the air.
b. False. For e.g. if the velocity of the particle is along the x-axis = V_{x} m/s, then its
speed = V_{x} m/s. Therefore if V_{x} = 0, then V_{x} = 0.
c. True.
d. False. It is true only if the speed is along the direction of motion.
Question-12
Solution:
First fall : Initial velocity v(0) = 0
Displacement x(0) â€“ x(t) = 90 m
v(t)^{2} â€“ v(0)^{2 }= 2 a [x(0) â€“ x(t)]
v(t)^{2 }= 2 Ã— 9.8 Ã— 90
v(t) = 42 m/s
v(t)= v(0) + a t
42 = 0 + 9.8 t
t = Time of fall = 4.3 s
First bounce: v(0)= 42 Ã— (-0.9) = -37.8 m/s
v(t) = 0 = - 37.8-9.8t
t = 3.86 s.
Question-13
a. Magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
b. Magnitude of average velocity over an interval of time and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater or equal to the first. When is the equality sign true? (For simplicity, consider one-dimensional motion only)?
Solution:
a. Magnitude of displacement considers only the starting and finishing points. The total length of path, on the other hand, considers the actual total movement of the particle during the time. For example, if a particle starts from A, goes to B in a straight line, and returns back to A in a straight line, the displacement is zero, whereas, the total path length = 2 ï‚´ï€ AB.
b. Average speed = total path length travelled / time taken
Average velocity = straight line distance between start and end points / time taken.
Since any path travelled cannot be shorter than the straight distance, the average speed cannot be less than the average velocity. They become equal only when the path actually travelled is the straight line between the start and end points.
Question-14
(a) magnitude of average velocity and
(b) average speed of the man over the interval of time
(i) 0 to 30 minutes (ii) 0 to 50 minutes (iii) 0 to 40 minutes?
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time and not as magnitude of average velocity. You would not like to tell the tired on his return home that his average speed was zero!]
Solution:
(a) The magnitude of average velocity=displacement/time taken
[The displacement is the straight line distance between the start and end points]
(b) (i) Average speed during 0 to 30 min =2.5 km /0.5 h=5 km/h
Average velocity during 0 to 30min = 2.5 km /0.5 h=5 km/h
(ii) During 0 to 50 min:
Distance travelled = 2.5 km+7.5 Ã— (20/60)km = 5 km
Average speed during 0 to 50 min = 5 km /(50/60) h = 6 km/h
Displacement = 2.5 - (7.5Ã— 20/60)km = 0 km
Average velocity during 0 to 50min = 0
(iii) During 0 to 40 min:
Distance travelled = 2.5 km + 7.5 Ã— (10/60)km = 3.75 km
Average speed during 0 to 40 min = 3.75 km /(40/60) h = 5.625 km/h
Displacement = 2.5 - (7.5 Ã— 10/60)km = 1.25 km
Average velocity during 0 to 40min = 1.25 km/(40/60) h = 1.875 km/h.
Question-15
Solution:
While evaluating instantaneous speed or magnitude of instantaneous velocity, we consider a very small displacement tending to zero (D x). Since it is very small, the length of path can be taken to be equal to displacement. This makes both the figures same.
Question-16
(d)
Solution:
(a) Not possibly represent a one-dimension motion, since the graph indicates two positions for each ordinate in time axis. This situation does not exist for one dimension motion.
(b) Not possibly represent a one-dimension motion, because a particle cannot have a positive and also a negative velocity at the same instance of time.
(c) Not possibly represent a one-dimension motion, because speed can never be negative. (Velocity can be).
(d) Not possibly represent a one-dimension motion, because total path length can never decrease with increase in time.
Question-17
Solution:
For t<0, there is no displacement with increase in time. The particle should have been stationary.
For t>0, the graph gives displacement and not the trajectory. There is increasing displacement in unit time (i.e. the velocity) when â€˜tâ€™ increases. This occurs in a particle subjected to uniform acceleration, for example, a body falling under gravity, starting from zero velocity at time zero.
Question-18
Solution:
[We know the ground velocities of the cars and relative velocity of the bulletw.r.to the police car.
They are to be related using the ground as the reference point.]
Velocity of the bullet w.r.to the police van= V_{bp}=150 m/s=540 km/h
Velocity of the policevan w.r.to the ground=V_{pg}=30 km/h
Velocity of the bullet w.r.to the ground=V_{bg}
V_{bg}= V_{bp}+V_{pg}
= 540 + 30km/h= 570 km/h
Velocity of the bullet w.r.to the thiefs car=V_{bt}
Velocity of the thief w.r.to the ground= V_{tg}=192 km/h
V_{bt}= V_{bg}+V_{gt}
= V_{bg}+(-V_{tg})
=570 -192km/h
= 378 km/h
= 105 m/s.
Question-19
(a)
(b)
(c)
Solution:
(a) A carom striker on a smooth board, which hits one side of the board perpendicularly, reflected with reduced velocity and falls into the pocket.
(b) A ball thrown vertically up , falling on the ground and rebounding with reduced speed.
(c) A ball with uniform speed is hit back with a bat, for very short interval of time.
Question-20
Solution:
Time |
X- Position |
V- Velocity |
a- Acceleration |
0.3s |
<0 |
<0 |
>0 |
1.2s |
>0 |
>0 |
<0 |
-1.2s |
<0 |
>0 |
>0 |
Question-21
Solution:
The average speed in a small interval of time is equal to the slope of the x-t graph in that interval. Since the slope of the graph is maximum in interval 3 and least in interval2, the speed is greatest in interval 3 and least in interval 2.
The slope of the x-t graph is positive in interval 1 and 2 and negative in interval 3. Therefore, speed is positive in intervals 1 and 2 and negative in interval 3.
Question-22
Solution:
(a) Average acceleration=(v_{2}- v_{1})/2 (1 unit time)=(v_{2}- v_{1})/2
considering only magnitude the average acceleration in the interval 2 is the greatest, though it is deceleration.
(b) Average speed is greatest in 3.
(c) Velocity is positive in all intervals.[While the magnitude of the speed increases and decreases, the velocity remains positive throughout. Hence the direction of motion does not change throughout. Considering the direction of motion
as the positive direction, the velocity remains positive.]
Acceleration is positive in the intervals 1 and 3 and negative in interval2.
(d) A, B, C and D are turning points of the magnitude of velocity and velocities remain constant in these instances. The acceleration in these points is therefore zero.
Question-23
Solution:
s_{nth}= v(0)+(9/2) (2n - 1)
x(0)= 0;v(0) = 0; a = 1 m/s^{2}; n =1,2,3,etc
Distance traveled in (n) s###ERROR###Ã¨ x(n) -x(0) = x(0) + Â½ an^{2}
Distance traveled in(n-1) s###ERROR###Ã¨ x(n-1) - x(0) = x(0) + Â½ a (n-1)^{2}
Distance traveled in n^{th }s###ERROR###Ã¨ x(n)- x(n-1)=Â½ a (2n-1)
In the given question distance covered during the n^{th }second = Â½(2n-1)a
n###ERROR###Ã¨ |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
x(n^{th})###ERROR###Ã¨ |
0.5 |
1.5 |
2.5 |
3.5 |
4.5 |
5.5 |
6.5 |
7.5 |
8.5 |
9.5 |
The graph is a straight line at an angle to the time axis.
Question-24
Solution:
When the lift is stationary, consider the ballâ€™s movement up.
v(0) = 49 m/s, v(t) = 0, a =-9.8 m/s^{2}, t = time in secs.
v(t) = v(0) + at
0 = 49 + (-9.8) t
t= 49 /9.8 = 5 s
Time for upward motion = t = 5 s
Time to return to start =2t = 10 s
When the lift starts moving up, the relative velocity of the ball with respect to the lift remains the same. The time to reach back does not alter relatively. Hence the time will again be only 10 s.
[Here the distance between the lift and the ball at any instant remains the same as it was at the corresponding instant in earlier case. Consider the height in case 1, as equivalent to the relative distance apart in case 2].
Question-25
Solution:
(a) For an observer outside when the child runs along the direction of the belt, the velocity = 9 + 4 = 13km/hr.
(b) For an observer outside when the child runs against the direction of the belt, the velocity = 9 - 4 = 5 km/hr.
(c) Time to run the distance of 50 m in either of the above cases will not depend on the velocity of the belt or the
direction of the run, since the child is moving relative to the belt and the parents are stationary relative to the belt.
Speed of the child = 9 km/h = 9000/ 3600 m/s = 2.5 m/s
Time taken = 50 / 2.5 = 20 sin either case.
For either of the parents the speed of the child is 9 m/sand the time taken is 20 s.
Question-26
Solution:
When a particle projected upward goes below the level of projection, it is easier to consider the total movement as one stretch and apply the formulas with due consideration for proper signs. It is also advantageous to consider the lowest point as datum so that the distance measurements are positive.
Consider the motion of the first stone.
x_{1}(0)=200 m,v_{1}(0)=15 m/s, a=- 10 m/s^{2}
x_{1}(t)=x_{1}(0) +v_{1}t++ Â½ a t^{2}
0=200 + 15 t_{1}- 5t_{1}^{2}
t_{1}=8 s
Similarly for the second stone,
x_{2}(0)=200 m,v_{2}= 30 m/s, a=10 m/s^{2}
0=200+30 t_{2}-5 t_{2}^{2}
t_{2}=10 s
[Further, during the period from the start till the first stone touches the ground at zero level, the relative velocity is 30 -15 m/s=15 m/s constant. This makes the relative displacement graph a straight line. After this time, the second stone, which is traveling downwards at 120 m/s continuously whereas the second stone is at, rest on the ground. The relative displacement follows the curve, which is the last portion of the displacement curve of the second stone alone.]
The graph is therefore verified.
Equations for the straight-line part:
Displacement=Relative velocityÃ— time=15 m/sÃ— t
Equation is x_{2}- x_{1}=15t_{1}
Equation for the parabolic portion is the same as that for the second stone.
i.e. x_{2}-x_{1}=-5 t_{1}^{2}+230 t_{1}+ 200.
Question-27
Solution:
[Distance traversed by a particle is given by the area bound by the velocity curve and the time axis, and the two ordinates.]
(a)Distance covered from t = 0s to t = 10s=Â½Ã— 10Ã— 12= 60m
Average speed= Distance travelled / time =60 / 10= 6 m/s
(b) Distance covered from t = 2s to 6s = Sum of the areas of the two trapeziums under the curve between the specified coordinates
={{(4.8 + 12) /2}Ã— 3} +{{(12 + 9.6) /2}Ã— 1}
=36 m.
Average speed= 36/4=9 m/s.
Question-28
(i) x (t_{2}) = x (t_{1}) + v (t_{1}) (t_{2}-t_{1}) + Â½ a (t_{2}-t_{1})^{2}
(ii) v(t_{2}) = v(t_{1}) + a (t_{2}-t_{1})
(iii)^{v}average =
(iv)^{a}average=
(v) x(t_{2}) = x (t_{1}) +^{v}average (t_{2}-t_{1}) + Â½^{a}average(t_{2}-t_{1})^{2}
(vi) x(t_{2}) - x(t_{1}) = area under the v-t curve bounded by the t-axis and the dotted line shown.
Solution:
(i) Wrong. There is an additional term (t_{2}- t_{1}).
(ii) Wrong. This is true only for constant acceleration.
(iii) Correct
(iv) Correct
(v) Wrong
(vi) Correct