**
Question.1: An
object
experiences a
net zero
external unbalancedforce.
Is it possible
for the object
to be travelling
with the
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__:
Yes, an object
may travel with
a non-zero
velocity even
when the net
external force
on it is zero. A
rain drop 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.2: When
a carpet is
beaten with a
stick, dust
comes out.
Explain, why?**

__
Solution__:
When a carpet is
beaten with a stick
it comes
into motion at
once. But the
dust particles
continue to be
at rest due to
inertia and get
detached from
the carpet.

**
Question.3: why
is it advised to
tie any luggage
kept on the
roof of
a bus with a
rope?**

__
Solution__:
Due to sudden
jerks or due to
the bus taking
sharp turns on
the road, the
luggage may fall
down from the
roof because
of its tendency
to continue to
be either at
rest or in
motion in the
same direction
(inertia of
motion). To
avoid this, it
advised to tie
the luggage kept
on the
roof of
a bus with a
rope.

**
Question.5: a
truck starts
from rest and
rolls down a
hill with
constant
acceleration. It
travels a
distance of 400
m in 20 sec.
Find its
acceleration.
Also find the
force acting on
it if its mass
is 7 metric
tones.**

__
Solution__:

Here, u = 0, s =
400 m, t = 20 s

We know, s = ut
+ ½ at^{2}

Or, 400 = 0 + ½
a (20)^{2}

Or, a = 2 ms^{–2}

Now, m = 7 MT =
7000 kg, a = 2
ms^{–2}

Or, F = ma =
7000 x 2 = 14000
N **Ans.**

**
Question.6: A
stone of 1 kg is
thrown with a
velocity of 20
ms**^{-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__:

Here, m = 1 kg,
u = 20 ms^{-1} v
= 0, s = 50 m

Since, v_{2} -
u_{2} =
2as,

Or, 0 - 20^{2} =
2a x 50,

Or, a = – 4 ms^{-2}

Force of
friction, F = ma
= – 4N *Ans*.

**
Question.7: An
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__:
Total mass, m =
mass of engine +
mass of wagons

Or, m = 8000 + 5
x 2000 = 18000
kg.

(a) The net
accelerating
force, F =
Engine force -
Frictional force

Or, F = 40000 -
5000 = 35000 N

(b) The
acceleration of
the train, a = F
÷ m = 35000 ÷
18000 = 1.94 ms^{–2}.

(c) The force of
wagon 1 on wagon
2

= The net
accelerating
force - (mass of
wagon x
acceleration)

= 35000 - (2000
x 1.94) =
31111.2 N *Ans*.

**
Question.10:
Using a
horizontal force
of 200 N, we
intend to move awooden
cabinet across
a floor at
constant
velocity. What
is the force of
friction that
will be exerted
on the cabinet? **

__
Solution__:
The cabinet will
move with
constant
velocity only
when the net
force on it is
zero.

Therefore, force
of friction on
the cabinet =
200 N, in a
direction
opposite to the
direction of
motion of the
cabinet.

**
Question.11: 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 ms**^{–1 }before
the collision
during which
they stick
together. What
will be the
velocity of the
combined object
after
collision?

__
Solution__:
Here, m_{1 }=
m_{2 }=
1.5 kg, u_{1} =
2.5 ms^{–1 }u_{2} =
–2.5 ms^{–1 }

Let v be the
velocity of the
combined object
after collision.
By the law of
conservation of
momentum,

Total momentum
after collision
= Total momentum
before
collision,

Or (m_{1} +
m_{2}) v
= m_{1}u_{1} +
m_{2}u_{2}

Or (1.5 + 1.5) v
= 1.5 x 2.5 +1.5
x (–2.5)
[negative sign
as moving in
opposite
direction]

Or v = 0 ms^{–1} **Ans**.

**
Question.12:
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__:
The logic is
that Action and
Reaction always
act on different
bodies, so they
can not cancel
each other. When
we push a
massive truck,
the force of
friction between
its tyres and
the road is very
large and so the
truck does not
move.

**
Question.13: A
hockey ball of
mass 200 gm
travelling at 10
ms**^{–1} is
struck by a
hockey stick so
as to
return it
along its
original path
with a velocity
at 5 ms^{–1}.
Calculate the
change of
momentum
occurred in the
motion of the
hockey ball by
the force
applied by the
hockey stick.

__
Solution__:

Change in
momentum = m (v
- u) = 0.2 (–5 -
10) = –3 kg ms^{–1}.

(The negative
sign indicates a
change in
direction of
hockey ball
after it is
struck by hockey
stick. Magnitude
of change in
momentum = 3 kg
ms^{–1}).

**
Question.16: An
Object of mass
100 kg is
accelerated
uniformly from a
velocity of 5 ms**^{–1} to
8 ms^{–1 }in
6 sec. Calculate
the initial and
final momentum
of the object.
Also find the
magnitude of the
force exerted on
the object.

__
Solution__:
Here, m = 100
kg, u = 5 ms^{–1},
v = 8 ms^{–1},
t = 6 sec.

Initial
momentum, p_{1} =
mu = 500 kg ms^{–1}

Final momentum,
p_{2 }=
mv = 800 kg ms^{–1 }

The magnitude of
the force
exerted on the
object, F = (p_{2} -
p_{1}) ÷
t = (800 - 500)
÷ 6 = 50 N *Ans*.