Momentum
Numertical Problems Based on Momentum
Type – I – (Calculation of Momentum)
Question:- 1 - What will
be the momentum of a stone having mass of 10 kg when it is thrown with a
velocity of 2m/s?
Solution:-
Given,
Mass (m) = 10kg
Velocity (v) = 2m/s
Momentum (p) =?
We know that, Momentum (p) = Mass
(m) x Velocity (v)
Therefore, p = 10kg x 2 m/s = 20
kg m/s
Thus the momentum of the stone =
20 kg m/s.
Question: 2 - Calculate
the momentum of a bullet of 25 g when it is fired from a gum with a velocity of
100m/s.
Solution:
Given,
Velocity of the bullet (v) =
100m/s
Mass of the bullet (m) = 25 g =
25/1000 kg = 0.025kg
Momentum (p) =?
Or, p = 2.5 kg m/s
Thus the momentum of the bullet =
2.5 kg m/s
Question: 3 - Calculate
the momentum of a bullet having mass of 25 g is thrown using hand with a
velocity of 0.1 m/s.
Solution:
Given,
Velocity of the bullet (v) =
0.1m/s
Mass of the bullet (m) = 25 g =
25/1000 kg = 0.025kg
Momentum (p) =?
We know that, Momentum (p) = Mass
(m) x Velocity (v)
Therefore, p = 0.025 kg x 0.1 m/s
Or, p = 0.0025 kg m/s
Thus the momentum of the bullet =
0.0025 kg m/s
Question:
4 - The mass of a goods lorry is 4000 kg and the mass of goods loaded on it is
20000 kg. If the lorry is moving with a velocity of 2m/s what will be its
momentum?
Solution:
Given,
Velocity
(v) = 2m/s
Mass of
lorry = 4000 kg, Mass of goods on the lorry = 20000 kg
Therefore,
total mass (m) of the lorry = 4000 kg + 20000 kg = 24000 kg
Momentum
(p) =?
We know
that, Momentum (p) = Mass (m) x Velocity (v)
Therefore,
p = 24000
kg x 2 m/s = 48000 kg m/s
Thus,
momentum of the lorry = 48000 kg m/s
Question:
5 - A car having mass of 1000 kg is moving with a velocity of 0.5m/s. What will
be its momentum?
Solution:
Given,
Velocity
of the bullet (v) = 0.5m/s
Mass of
the bullet (m) = 1000 kg
Momentum
(p) =?
We know
that, Momentum (p) = Mass (m) x Velocity (v)
Therefore,
p = 1000 kg x 0.5 m/s
Or, p =
500 kg m/s
Thus the
momentum of the bullet = 500 kg m/s
Type – II (Calculation of Mass)
Question:
1 – A vehicle is running with a velocity of 5m/s. If the momentum of the
vehicle is 5000 kg m/s, what is its mass?
Solution:
Given,
Momentum
(p) = 5000 kg m/s
Velocity
(v) = 5m/s
Mass (m)
=?
We know
that, Momentum (p) = Mass (m) x Velocity (v)
Therefore,
5000 kg m/s = m × 5m/s
Thus,
mass of the vehicle = 1000 kg
Question:
2 – A stone attains a momentum of 1 kg m/s when it flies with a velocity of
2m/s, then what will be mass of the stone?
Solution:
Given,
Momentum
(p) = 1 kg m/s
Velocity
(v) = 2m/s
Mass (m)
=?
We know
that, Momentum (p) = Mass (m) x Velocity (v)
Therefore,
1 kg m/s = m × 2m/s
Thus,
mass of the stone = 0.5 kg or 500 g
Question:
3 – When a bullet is fired from a rifle its momentum become 20 kg m/s. If the
velocity of the bullet is 1000m/s what will be its mass?
Solution:
Given,
Momentum
(p) = 20 kg m/s
Velocity
(v) = 1000m/s
Mass (m)
=?
We know
that, Momentum (p) = Mass (m) x Velocity (v)
Therefore,
20 kg m/s = m × 1000m/s
Thus,
mass of the bullet = 20 g
Question:
4 – When a missile is fired from a tank it gets a momentum of 2000 kg m/s. If
the velocity of the missile is 50m/s what will be its mass?
Solution:
Given,
Momentum
(p) = 2000 kg m/s
Velocity
(v) = 50m/s
Mass (m)
=?
We know
that, Momentum (p) = Mass (m) x Velocity (v)
Therefore,
2000 kg m/s = m × 50 m/s
Thus,
mass of the missile = 40 g
Question:
5 – A bird is flying with a velocity of 3 m/s. If the momentum of the bird is
3.60 kg m/s what is its mass?
Solution:
Given,
Momentum
(p) = 3.60 kg m/s
Velocity
(v) = 3 m/s
Mass (m)
=?
We know
that, Momentum (p) = Mass (m) x Velocity (v)
Therefore,
3.60 kg m/s = m × 3 m/s
Thus,
mass of the bird = 1 kg 200 g
Type – III (Calculation of velocity)
Question:
1 – If the momentum of a flying brick is 50 kg m/s and its mass is 10 kg.
Calculate its velocity?
Solution:
Given,
Momentum
(m) = 50kg m/s
Mass (m)
= 10kg
Velocity
(v) =?
We know
that,
Momentum
(p) = Mass (m) x Velocity (v)
⇒ 50 kg m/s = 10kg × v
Thus,
velocity of the brick = 5m/s
Question:
2 – A bullet of 25 g is when fired from a piston gets a momentum of 50 kg m/s.
Calculate the velocity of bullet.
Solution:
Given,
Momentum
(m) = 50kg m/s
Mass (m)
= 25 g = 25/1000 kg = 0.025 kg
Velocity
(v) =?
We know
that,
Momentum
(p) = Mass (m) x Velocity (v)
⇒ 50 kg m/s = 0.025 kg × v
Thus,
velocity of the bullet = 2000 m/s
Question:
3 – A vulture when flying with a velocity ‘v’ attains a momentum of 20 kg m/s.
If the mass of the vulture is 25 kg what is the value of ‘v’?
Solution:
Given,
Momentum
(m) = 20 kg m/s
Mass (m)
= 25 kg
Velocity
(v) =?
We know
that, Momentum (p) = Mass (m) x Velocity (v)
⇒ 20 kg m/s = 25 kg × v
Thus,
velocity of the vulture = 8 m/s
Question:
4 – A brick after falling from a hill collide with ground with a momentum of
100 kg m/s. If the mass of the brick is 5 kg what was its velocity while
colliding with the ground?
Solution:
Given,
Momentum
(m) = 100kg m/s
Mass (m)
= 5kg
Velocity
(v) =?
We know
that,
Momentum
(p) = Mass (m) x Velocity (v)
⇒ 100 kg m/s = 5kg × v
Thus,
velocity of the brick = 20 m/s
Question:
5 – Calculate the velocity of a missile having mass of 100 kg, if it attains a
momentum of 5000 kg m/s when fired from a rocket gun?
Solution:
Given,
Momentum
(m) = 5000kg m/s
Mass (m)
= 100kg
Velocity
(v) =?
We know
that, Momentum (p) = Mass (m) x Velocity (v)
⇒ 5000 kg m/s = 100kg × v
Thus,
velocity of the missile = 50 m/s
NCERT In Text Solution
Question:- 1 - Which of
the following has more inertia:
(a) a rubber ball and a
stone of the same size?
(b) a bicycle and a
train?
(c) a five rupees coin
and a one-rupee coin?
Answer:-
(a) A stone.
(b) A train
(c) A five rupees coin.
Explanation – Inertia is
associated with mass. Objects having more mass have more inertia.
Question:- 2 - In the
following example, try to identify the number of times the velocity of the ball
changes:
“A football player kicks
a football to another player of his team who kicks the football towards the
goal. The goalkeeper of the opposite team collects the football and kicks it
towards a player of his own team”. Also identify the agent supplying the force
in each case.
Answer:
The velocity of football
changes four times.
First, when a football
player kicks to another player, second when that player kicks the football to
the goalkeeper. Third when the goalkeeper stops the football. Fourth when the
goalkeeper kicks the football towards a player of his own team.
Agent supplying the
force –
First case – First
player
Second case – Second
player
Third case – Goalkeeper
Fourth case – Goalkeeper
Question: 3 - Explain
why some of the leaves may get detached from a tree if we vigorously shake its
branch.
Answer: The answer of
this cause lies behind the Newton’s First Law of Motion. Initially, leaves and
tree both are in rest. But when the tree is shaken vigorously, tree comes in
motion while leaves have tendency to be in rest. Thus, because of remaining in
the position of rest some of the leaves may get detached from a tree if we
vigorously shake its branch.
Question:
4 - Why do you fall in the forward direction when a moving bus brakes to a stop
and fall backwards when it accelerates from rest?
Answer:
In
a moving bus, passengers are in motion along with bus. When brakes are applied
to stop a moving bus, bus comes in the position of rest. But because of
tendency to be in the motion a person falls in forward direction.
Similarly,
when a bus is accelerated from rest, the tendency to be in rest, a person in
the bus falls backwards.
Question:
5 - If action is always equal to the reaction, explain how a horse can pull a
cart.
Answer:
Horse pushes the ground in backward direction. In reaction to this, the horse
moves forward and cart moves along with horse in forward direction.
In
this case when horse tries to pull the cart in forward direction, cart pulls
the horse in backward direction, but because of the unbalanced force applied by
the horse, it pulls the cart in forward direction.
Question:
6 - Explain, why is it difficult for a fireman to hold a hose, which ejects
large amounts of water at a high velocity.
Answer:
When large amount of water is ejected from a hose at a high velocity, according
to Newton’s Third Law of Motion, water pushes the hose in backward direction
with the same force. Therefore, it is difficult for a fireman to hold a hose in
which ejects large amount of water at a high velocity.
Question:
7 - From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial
velocity of 35 m/s. Calculate the initial recoil velocity of the rifle.
Answer:
Given,
Mass
of rifle (m1) = 4 kg
Initial
velocity of rifle (u1) = 0
Mass
of bullet (m2) = 50 g = 50/1000 kg = 0.05 kg
Initial
velocity of bullet (u2) = 0
Final
velocity (v2) of bullet = 35 m/s
Final
velocity [Recoil velocity] of rifle (v1)=?
We
know that,
Here
negative sign of velocity of rifle shows that rifle moves in the opposite
direction of the movement of bullet. Therefore, recoil velocity of rifle is
equal to 0.44 m/s.
Question:
8 - Two objects of masses 100 g and 200 g are moving along the same line and
direction with velocities of 2 m/s and 1 m/s, respectively. They collide and
after the collision, the first object moves at a velocity of 1.67 m/s.
Determine the velocity of the second object.
Answer:
Given,
Mass
of first object (m1) = 100 g = 100/1000 kg = 0.1 kg
Initial
velocity of first object (u1) = 2 m/s
Final
velocity of first object after collision (v1) = 1.67 m/s
Mass
of second object (m2) = 200 g = 200/1000 kg = 0.2 kg
Initial
velocity of second object (u2) = 1 m/s
Final
velocity of second object after collision (v2) =?
We
know that,
Question: 1 - An object
experiences a net zero external unbalanced force. Is it possible for the object
to be travelling with a 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.
Answer: When a net zero
external unbalanced force is applied on the body, it is possible for the object
to be travelling with a non-zero velocity. In fact, once an object comes into
motion and there is a condition in which its motion is unopposed by any
external force; the object will continue to remain in motion. It is necessary
that the object moves at a constant velocity and in a particular direction.
Question: 2 - When a
carpet is beaten with a stick, dust comes out of it. Explain.
Answer: Beating of a
carpet with a stick; makes the carpet come in motion suddenly, while dust
particles trapped within the pores of carpet have tendency to remain in rest,
and in order to maintain the position of rest they come out of carpet. This
happens because of the application of Newton’s First Law of Motion which states
that any object remains in its state unless any external force is applied over
it.
Question:- 3 - Why is it
advised to tie any luggage kept on the roof of a bus with a rope?
Answer:- Luggage kept on
the roof of a bus has the tendency to maintain its state of rest when bus is in
rest and to maintain the state of motion when bus is in motion according to
Newton’s First Law of Motion.
When bus will come in
motion from its state of rest, in order to maintain the position of rest,
luggage kept over its roof may fall down. Similarly, when a moving bus will
come in the state of rest or there is any sudden change in velocity because of
applying of brake, luggage may fall down because of its tendency to remain in
the state of motion.
This is the cause that
it is advised to tie any luggage kept on the roof a bus with a rope so that
luggage can be prevented from falling down.
Question: 4 - A batsman
hits a cricket ball which then rolls on a level ground. After covering a short
distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not
hit the ball hard enough.
(b) velocity is
proportional to the force exerted on the ball.
(c) there is a force on
the ball opposing the motion.
(d) there is no
unbalanced force on the ball, so the ball would want to come to rest.
Answer: (c) There is a
force on the ball opposing the motion.
Explanation: When ball
moves on the ground, the force of friction opposes its movement and after some
time ball comes to the state of rest.
Question:
5 - A truck starts from rest and rolls down a hill with a constant
acceleration. It travels a distance of 400 m in 20 s. Find its acceleration.
Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)
Answer:
Given,
Initial
velocity of truck (u) =0 (Since, truck starts from rest)
Distance
travelled, s = 400 m
Time
(t) = 20 s
Acceleration
(a) = ?
We
know that,
Force
acting upon truck:
Given
mass of truck = 7 ton = 7 X 1000 kg = 7000 kg
We
know that, force, P = m x a
Therefore,
P = 7000 kg x 2 m s - 2
Or,
P = 14000 Newton
Thus,
Acceleration = 2 m s - 2 and force acting upon truck in the given condition =
14000 N
Question:
6 - A stone of 1 kg is thrown with a velocity of 20 m/s 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?
Answer:
Given,
Mass
of stone = 1 kg
Initial
velocity, u = 20 m/s
Final
velocity, v = 0 (as stone comes to rest)
Distance
covered, s = 50 m
Force
of friction =?
We
know that,
Now,
we know that, force, F = mass x acceleration
Therefore,
F = 1 kg X -4ms-2
Or,
F = -4ms-2
Thus,
force of friction acting upon stone = -4ms-2. Here negative sign shows that
force is being applied in the opposite direction of the movement of the stone.
Question: 7 - A 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.
Answer:-
Given, force of engine =
40000 N
Force of friction = 5000
N
Mass of engine = 8000 kg
Total weight of wagons =
5 x 2000 kg = 10000 kg
(a) The net accelerating
force
= Force exerted by engine
– Force of fricition
= 40000 N – 5000 N =
35000 N
(b) The acceleration of
the train
We know that, F = mass x
acceleration
Or, 35000 N = (mass of
engine + mass of 5 wagons) X a
Or, 35000 N = (8000 kg +
10000 kg) X a
Or, 35000N = 18000 kg X
a
(c) The force of wagon 1
on wagon 2
Since, net accelerating
force = 35000 N
Mass of all 5 wagons =
10000 kg
We know that, F = m x a
Therefore, 35000N =
10000 kg x a
Question:- 8 - An
automobile vehicle has a mass of 1500 kg. What must be the force between the
vehicle and road if the vehicle is to be stopped with a negative acceleration
of 1.7 m s-2?
Answer:
Given,
Mass of the vehicle, m =
1500 kg
Acceleration, a = - 1.7
m s -2
Force acting between the
vehicle and road, F =?
We know that, F = m x a
Therefore, F = 1500 kg X
1.7 m s-2
Or, F = - 2550 N
Thus, force between
vehicle and road = - 2550 N. Negative sign shows that force is acting in the
opposite direction of the vehicle.
Question:
9 - What is the momentum of an object of mass m, moving with a velocity v?
(a)
(mv)2 (b) mv2 ( c ) Ω mv2 (d) mv
Answer:
(d) mv
Explanation:
Given,
mass = m, velocity = v, therefore, momentum =?
We
know that, momentum, P = mass x velocity
Therefore,
P = mv
Thus,
option (d) mv is correct
Question:
10 - Using a horizontal force of 200 N, we intend to move a wooden cabinet
across a floor at a constant velocity. What is the friction force that will be
exerted on the cabinet?
Answer:
Since,
a horizontal force of 200N is used to move a wooden cabinet, thus a friction
force of 200N will be exerted on the cabinet. Because according to third law of
motion, an equal magnitude of force will be applied in the opposite direction.
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 m s-1 before the
collision during which they stick together. What will be the velocity of the
combined object after collision?
Answer:
Since, two objects of equal mass are moving in opposite direction with equal
velocity, therefore, the velocity of the objects after collision during which
they stick together will be zero.
Explanation:
Given,
Mass
of first object, m1 = 1.5 kg
Mass
of second object, m2 = 1.5 kg
Initial
velocity of one object, u1 = 2.5 m/s
Initial
velocity of second object, u2 = -2.5 m/s (Since second object is moving in
opposite direction)
Final
velocity of both the objects, which stick after collision, v =?
We
know that,
Therefore,
final velocity of both the objects after collision will be zero.
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.
Answer:
Because of the huge mass of the truck, the force of static friction is very
high. The force applied by the student is unable to overcome the static
friction and hence he is unable to move the truck. In this case, the net
unbalanced force in either direction is zero which is the reason of no motion
happening here. The force applied by the student and the force because of
static friction are cancelling out each other. Hence, the rationale given by
the student is correct.
Question:
13 - A hockey ball of mass 200 g travelling at 10 m/s is struck by a hockey
stick so as to return it along its original path with a velocity at 5 m/s.
Calculate the change of momentum occurred in the motion of the hockey ball by
the force applied by the hockey stick.
Answer:
Given,
Mass
of hockey ball, m = 200 g = 200/1000 kg = 0.2 kg
Initial
velocity of hockey ball, u = 10 m/s
Final
velocity of hockey ball, v = - 5 m/s (because direction becomes opposite)
Change
in momentum =?
We
know that,
Momentum
= mass x velocity
Therefore,
Momentum of ball before getting struck = 0.2 kg x 10 m/s = 2 kg m/s
Momentum
of ball after getting struck = 0.2 kg x - 5m/s = - 1 kg m/s
Therefore,
change in momentum = momentum before getting struck – momentum after getting
struck
=
2 kg m/s – (-1 kg m/s) = 2 kg m/s + 1 kg m/s = 3 kg m/s
Thus,
change of momentum of ball after getting struck = 3 kg m/s
Question:- 14 - A bullet
of mass 10 g travelling horizontally with a velocity of 150 m/s strikes a
stationary wooden block and comes to rest in 0.03 s. Calculate the distance of
penetration of the bullet into the block. Also calculate the magnitude of the
force exerted by the wooden block on the bullet.
Answer:
Given,
Mass of bullet, m = 10 g
= 10/1000 kg = 0.01 kg
Initial velocity of
bullet, u = 150 m/s
Since bullet comes to
rest, thus final velocity, v =0
Time, t = 0.03 s
Distance of penetration,
i.e. Distance, covered (s)=?
Magnitude of force
exerted by wooden block =?
We know that,
Magnitude of force
exerted by wooden block
We know that, Force =
mass x acceleration
Or, F = 0.01 kg x – 5000
m s-2 = - 50 N
Therefore,
Penetration of bullet in
wooden block = 2.25 m
Force exerted by wooden
block on bullet = - 50 N. Here negative sign shows that force is exerted in the
opposite direction of bullet.
Question: 15 - An object
of mass 1 kg travelling in a straight line with a velocity of 10 m/s collides
with, and sticks to, a stationary wooden block of mass 5 kg. Then they both
move off together in the same straight line. Calculate the total momentum just
before the impact and just after the impact. Also, calculate the velocity of
the combined object.
Answer:
Given, mass of moving
object, m1 = 1 kg
Mass of the wooden
block, m2 = 5kg
Initial velocity of
object, u1 = 10 m/s
Initial velocity of
wooden block, u2 = 0
Final velocity or moving
object and wooden block, v =?
Total momentum before
collision and after collision =?
We know that,
Thus,
Velocity of both the
object after collision = 1.66 m/s
Total momentum before
collision = 10 kg m/s
Total momentum after
collision = 10 kg m/s
Question:
16 - An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s
to 8 m/s in 6 s. Calculate the initial and final momentum of the object. Also,
find the magnitude of the force exerted on the object.
Answer:
Given,
Initial
velocity, u = 5 m/s
Final
velocity, v = 8 m/s
Mass
of the given object, m = 100 kg
Time,
t = 6 s
Initial
momentum and Final momentum =?
Magnitude
of force exerted on the object =?
We
know that,
Momentum
= mass x velocity
Therefore,
initial momentum = mass x initial velocity
=
100 kg X 5 m/s = 500 kg m/s
Final
momentum = mass x final velocity
=
100 kg x 8 m/s = 800 kg m/s
We
know that,
Now,
Force exerted on object = Mass x Acceleration
=
100 kg 0.5 m/s/s
=
50 N
Question:
17 - Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a
high velocity on an expressway when an insect hit the windshield and got stuck
on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran
suggested that the insect suffered a greater change in momentum as compared to
the change in momentum of the motorcar (because the change in the velocity of
the insect was much more than that of the motorcar). Akhtar said that since the
motorcar was moving with a larger velocity, it exerted a larger force on the
insect. And as a result the insect died. Rahul while putting an entirely new
explanation said that both the motorcar and the insect experienced the same
force and a change in their momentum. Comment on these suggestions.
Answer:
We know, that as per the Law of Conservation of Momentum; total momentum of a
system before collision is equal to the total momentum of the system after collision.
In
this case, since the insect experiences a greater change in its velocity so it
experiences a greater change in its momentum. From this angle, Kiran’s
observation is correct.
Motorcar
is moving with a larger velocity and has a bigger mass; as compared to the
insect. Moreover, the motorcar continues to move in the same direction even
after the collision; which suggests that motorcar experiences minimal change in
its momentum, while the insect experiences the maximum change in its momentum.
Hence, Akhtar’s observation is also correct.
Rahul’s
observation is also correct; because the momentum gained by the insect is equal
to the momentum lost by the motorcar. This also happens in accordance to the
law of conservation of momentum.
Question:
18 - How much momentum will a dumb-bell of mass 10 kg transfer to the floor if
it falls from a height of 80 cm? Take its downward acceleration to be 10 m/s-2.
Answer:
Given,
Mass
of dumb-bell = 10 kg
Distance,
s = 80 cm = 80/100 = 0.8 m
Acceleration,
a = 10 m/s/s
Initial
velocity of dumb-bell, u = 0
Momentum
=?
We
know that,
Now,
we know that, momentum = mass x velocity
=
10 kg x 4 m/s = 40 kg m/s
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