Force and Laws of Motion
CBSE Board - Class IX, Science (Physics)
Solutions of NCERT (CBSE) Science Textbook Exercise Questions
Question.1: An object experiences
a net zero external unbalanced force. 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 at2
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, v2
- u2 = 2as,
Or, 0 - 202
= 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 a wooden 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, m1
= m2 = 1.5 kg, u1 = 2.5 ms–1 u2
= –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 (m1
+ m2) v = m1u1 + m2u2
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, p1 = mu = 500 kg ms–1
Final
momentum, p2 = mv = 800 kg ms–1
The
magnitude of the force exerted on the object, F = (p2 - p1)
÷ t = (800 - 500) ÷ 6 = 50 N Ans.
Class IX, FORCE AND LAWS OF MOTION - must read
- Class 9 Ncert Cbse Guide - Science Chapter 9, Force and Laws of Motion - Important sample questions [Read]
- Force and Laws of Motion - Class 9 Ncert Cbse Science
Physics - MCQs [Read]
- Class IX, Solutions of Ncert Cbse Science Physics - Lesson 9, Force and Laws of Motion - Chapter In-text Questions [Read]
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