Class 9 science chapter 9 examples from S Chand.
Please send the pics of the examples based on numericals.
Answers
Question 1
How much net force is required to accelerate a 1000 kg car at 4.00 m/s2?
Solution
F
=
m
a
Given a=4.00 m/s2
m=1000kg
Therefore ,
F
=
m
a
=
1000
×
4
=
4000
N
Question 2
If you apply a net force of 3 N on .1 kg-box, what is the acceleration of the box
(a) 2 m/s2
(b) 30 m/s2
(c) 10 m/s2
(d) None of these.
Solution
Given F=3 N
m=.1 kg
F
=
m
a
or
a
=
F
m
=
30
m/s2
Question 3
A body of mass 1 kg undergoes a change of velocity of 4m/s in 4s what is the force acting on it?
Solution
Given
Δ
v
=
4
m
/
s
,t=4 s ,m=1kg
Acceleration is given by
a
=
Δ
v
t
a=1 m/s2
Now force is given by
F
=
m
a
F=1 N
Question 4.
A particle of 10 kg is moving in a constant acceleration 2m/s2 starting from rest. What is its momentum and velocity per the table given below
S.No
time
Momentum
Velocity
1
1sec
2
1.5 sec
3
2 sec
4
2.5 sec
Solution
Velocity can find using
v
=
u
+
a
t
For u=0
v
=
a
t
Momentum
P
=
m
v
S.No
time
Momentum
Velocity
1
1sec
20 Kg m/s
2 m/
2
1.5 sec
30 kg m/s
3 m/s
3
2 sec
40 kg m/s
4 m/s
4
2.5 sec
50 kg m/s
5 m/s
Question 5
If a net force of 7 N was constantly applied on 400 g object at rest, how long will it take to raise its velocity to 80 m/s?
a. 0 s
b. 2.23 s
c. 3.47 s
d. 4.57 s
Solution
Given F=7 N,m=400g= .4 kg Acceleration is given by
a
=
F
m
a=17.5 m/s2
Now u=0,v=80 m/s
v
=
u
+
a
t
t
=
v
−
u
a
t=4.57 sec
Question 6
A sedan car of mass 200kg is moving with a certain velocity . It is brought to rest by the application of brakes, within a distance of 20m when the average resistance being offered to it is 500N.What was the velocity of the motor car?
Solution
F
=
m
a
or
a
=
F
m
or
a= -500/200=-2.5 m/s2
Now
v
2
=
u
2
+
2
a
s
Now v=0,s=20 m,a=-2.5 m/s2
So, u=10 m/s
Question 7
A driver accelerates his car first at the rate of 4 m/s2 and then at the rate of 8 m/s2 .Calculate the ration of the forces exerted by the engines?
Solution
F
1
=
m
a
1
and
F
2
=
m
a
2
So,Ratio of force exerted is given by
=
F
1
F
2
=
m
a
1
m
a
2
=
a
1
a
2
=
1
:
2
Question 8
An object of mass 10 g is sliding with a constant velocity of 2 m/ s on a frictionless horizontal table. The force required to keep the object moving with the same velocity is
(a) 0 N
(b) 5 N
(c) 10 N
(d)20 N
Solution
As m=0, F=0
Hence (a) is correct
Question 9
A cricket ball of mass 0.20 kg is moving with a velocity of 1.2m/s . Find the impulse on the ball and average force applied by the player if he is able to stop the ball in 0.10s?
Solution
Impulse= Change in momentum
I
=
Δ
p
=
m
Δ
v
=
.20
×
1.2
=
.12
Kgm/s
Now
Impulse is also defined as
I
=
F
×
t
or
F
×
t
=
.12
or
F
=
.12
.10
=
1.2
N