Question-1
Solution:
We know that:
Momentum=Mass x velocity = m x v
Here, Mass, m = 75Kg
And, Velocity, v = 2m/s
Putting these values in the above formula, we get:
Momentum = 75 x 2Kg.m/s
= 150 kg .m/s.
Question-2
Solution:
Initial speed, u = 10m/s
Final speed, v = 0 (Scooter stops)
Acceleration, a = -0.5m/s^{2}
And, Distance covered, s = ?(To be calculated)
Now, putting these values in the third equation of motion:V^{2 }= u^{2 }+ 2as
We get, (0)^{2 }= (10)^{2 }+ 2 x (-0.5) x s
0 = 100-s
s = 100m
Thus, the distance covered in 100 metres.
Question-3
(i) What is the total distance travelled?
(ii) What is his resultant displacement?
Solution:
(i) Total distance travelled = 1.5+2.0+4.5 = 8.0m
(ii)To find the resultant displacement we should draw a map of the manâ€™s movements by choosing a convenient scale. Let 1cm represent 1m. Then 1.5m can be represented by 1.5cm long line,2.0m by 2.0cm line and 4.5m by a 4.5cm long line.
Now, 1cm = 1m
So, 6.3 cm = 6.3m
We draw a 1.5cm long line AB from west to east to represent 1.5m towards east. Then we draw a 2.0cm long line BC towards south to represent 2.0m towards south. And finally we draw a third line CD,4.5cm long, towards east to represent a distance of 4.5m towards east. Now, the resultant displacement can be found by joining the starting point A with the finishing point D. Thus, the line AD represents the final displacement of the man. Let us measure the length of line AD. It is found to be 6.3cm.
Thus, the final displacement as represented by AD is 6.3metres.
So, whenever a body travels along a zigzag, the final displacement is obtained by joining the starting point and the finishing point of the body by a straight line.
Question-4
Solution:
The resultant displacement can be given using the Pythagoras theorem
Resultant R ===10cm.
Question-5
Solution:
Here the term speed is used in the same sense as velocity.
Initial speed or velocity, u=56km/h =m/s = 15.55m/s
Final speed or velocity, v,= 58km/h = m/s = 16.11m/s
Acceleration = a=? (to be calculated)
Time, t = 1s
Now, putting these values in the first equation of motion:
V=u +at
We get:16.11 = 15.55 + a x 1
a=16.11-15.55 a = 0.56m/s^{2}
Thus, the acceleration of the ship is 0.56m/s^{2}.
Question-6
Solution:
Initial speed, u=20m/s
Final speed, v =?
Acceleration, a = 4m/s^{2}
And, Time, t = 2s
Now, v = u + at
So, V = 20 + 4 x 2
V = 20 + 8
V = 28m/s
Thus, after two seconds, the speed will be 28m/s.
Question-7
Solution:
Initial speed, u = 20km/h =m/s = 5.5m/s
Final speed, v = 50km/h =m/s = 13.8m/s
Acceleration, a = ? (to be calculated) And, Time t=10s V = u +at
13.8 = 5.5 + a x 10
10a = 13.8 - 5.5
10a = 8.3
a = 8.3/10
= 0.83m/s^{2}
Thus, the acceleration is 0.83m/s^{2}.
Question-8
(i) Speed
(ii) Acceleration.
Solution:
The two quantities, the slope of whose graph gives speed is
(i) Distance- Time.
And that of Acceleration is (ii) Speed-time.
Question-9
Solution:
The motion of a body whose distance-time graph is a straight line parallel to the time axis indicates that the body is not moving, it is stationary.
Question-10
Solution:
The average speed of a body is the total distance travelled divided by the total time taken to cover this distance. The speed of a body is usually not constant and the distance travelled divided by time gives us the average speed during that time. For example, a car which travels a distance of 100km in 4hours, the average speed is = 25km per hour. Although the average speed of the car is 25km per hour, it does not mean that the car is moving at this speed all the time. When the road is straight, flat and free, the speed may be much more than 25km per hour but on bends (curved road), hills or in a crowded area, the speed may fall well below this average value. We should remember that:
Average speed=.
Question-11
Solution:
S.No. |
Displacement |
Distance |
1. |
The shortest distance measured from the initial to the final position of a body is known as displacement. |
The total distance covered by a body from the initial position to arrive at the final position is called distance. |
2. |
Displacement of a body specifies the direction along which the distance is to be measured. |
It does not specify the direction. |
3. |
It is a vector quantity. |
It is a scalar quantity. |
Question-12
Solution:
The velocity of a body is said to be uniform when it covers equal distances in equal intervals of time however small the intervals may be, the direction remaining constant.
The velocity of a body is said to be non-uniform when it passes over unequal distances in successive equal intervals of time or changes its direction of motion.
For example, a train has a uniform velocity of 6m per second if it regularly moves a distance of 6m in one second in a constant direction. If the train does not regularly traverse a distance of 6m in a second, but instead halts at station, then its velocity at the time of leaving a station or reaching it will be different from its velocity between the station and so we state that it has non-uniform velocity.
Question-13
Overs |
Runs scored | |||||||||
Team A | 1 | 2 | 3 | 4 | 5 | 2 | 4 | 1 | 0 | 6 |
Team B | 1 | 2 | 3 | 4 | 5 | 3 | 10 | 4 | 5 | 3 |
Solution:
Question-14
Solution:
(i) We can determine the position of the car at any intermediate time with the help of its distance-time graph.
(ii) We can determine the location of the train at any given time with the help of its distance-time graph.
(iii) We can compute the velocity of a car from its distance-time graph.
(iv) We can compute the distance moved by a body in a given time from its velocity-time graph.
Question-15
Solution:
Let us consider a body moving in uniform motion, then we can plot the velocity-time graph for the motion of the body moving with a constant velocity of 30 km/h, which means that the body will cover a distance of 30 km in1 hour, 60km in 2 hours, 90 km in 3 hours and so on.
If we want to know the distance the body has moved between time say t_{1} and t_{2}. We draw perpendiculars from points corresponding to the time t_{1} and t_{2} on the above graph.
The two perpendiculars enclose a rectangle ABCD between the graph and the time axis. The sides AB and CD of this rectangle are equal to (t_{2} â€“ t_{1}) while the sides AC and BD are equal to 30 km/h. We know that the distance s covered by a body moving a velocity v, in time t is given by
s = vt
Hence, the distance moved by the body in time (t_{2} â€“ t_{1}) is
= [(30 km/h)(t_{2} â€“ t_{1})h]
= 30 (t_{2} â€“ t_{1}) km
The area under the velocity-time graph gives the distance travelled by the body.
Question-16
Time in s | Distance traveled in m |
0 | 0 |
2 | 1 |
4 | 4 |
6 | 9 |
8 | 16 |
10 | 25 |
12 | 36 |
Find out the motion of the body.
Solution:
on of the body is uniform acceleration motion.
Question-17
Solution:
Let us consider a body whose initial velocity is u, moves for a time t with a constant acceleration a, and so acquires a final velocity v. It moves a distance s in time t. Now, from the definition of acceleration, the change in velocity = v â€“ u
Therefore the rate of change of velocity = acceleration a =
or v â€“ u = at ----(1)
v = u + at ----(2)
Since distance = average Ã— velocity Ã— time
s = ----(3)
Substituting the value of v from equation (2), we get
s =
=
s = ut + at^{2} ----(4)
From the equation (1), we have
v â€“ u = at
and from (3) we have
2s = (u + v) Ã— t
u + v = ----(5)
Multiplying (1) by (5), we get
(v â€“ u) (v + u) = at Ã—
v^{2} â€“ u^{2} = 2as ----(6)
The equations (2), (4) and (6) are the equations of motion.
Question-18
Solution:
Initial velocity u = 36 km/h
= 36 Ã— m/s
= 10 m/s
Final velocity v = 0
Distance s = 100m
From the equations of motion, we have
v^{2} â€“ u^{2} = 2as
a =
=
= = -0.5 m/s.
Question-19
Solution:
Question-20
Solution:
If an object moves in a circle at a constant speed, it is said to be in uniform circular motion.
Examples:
(i) The moon revolving round the earth,
(ii) A satellite in circular orbit round a planet and
(iii) A cyclist moving at a constant speed on a circular track.