Class 11 Medical Physics - Laws Of Motion

Coefficient of static friction is depends on _______________

Static Friction:-For an object at rest on a flat table, Static friction is zero...The maximum static friction force is$$(F_s)_max=\mu_s N$$where $$\mu_s$$is the coefficient of static friction. Static friction is subtle because the static friction force is variable and depends on the external forces acting on an object.

Fill in the blanks:
Friction is a type of ________.

Contact force.

Find the moment of force of $$ 20 N $$ about an axis of rotation at distance $$ 0.5 $$ m from the force.

$$ F = 20 N \quad \bot Distance = 0.5 m $$

Moment of force = force $$ \times $$ perpendicular distance of force from the point of rotation.

$$ = F \times \bot distance $$

$$ = 20 N \times 0.5 m $$

$$ = ION m $$

Which are the following occation friction is beneficial and non beneficial
striking a match sticks on the side of a match box
wear and tear of machines.
ability to hold objects
wearing out of tyres
walking $$6.$$ loss of fuel

Beneficial :$$1,3,5$$
non beneficial :$$2,4,6$$

Acceleration of block A and B.

F.B.D. of 50 kg
$$N_2\,=\, 50\,g\, =\, 500\, N$$
Along horizontal direction, there is no force $$a_b\,=\, 0$$

A hockey ball of mass $$200$$g travelling at $$10$$m $$s^{-1}$$ is struck by a hockey stick so as to return it along its original path with a velocity at $$5$$m $$s^{-1}$$. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Initial velocity of the ball, $$u = -10\ ms^{-1}$$ (- sign shows the opposite direction)

Final velocity of the ball, $$v = 5$$ $$ms^{-1}$$

Mass of the ball, $$m = 200\ g = 0.2\ kg$$

Net change in the momentum of the ball, $$\Delta P = P_f-P_i$$

$$\Delta P = mv- mu = m \times 5 - m \times (-10) $$

$$\therefore \Delta P = 15 \times m = 15 \times 0.2 = 3\ {kg\ m\ s}^{-1}$$

A block weighing 20N rests on a horizontal surface. The coefficient of static friction between the block and surface is 0.4 and the coefficient of kinetic friction is 0.20.
a. How much is the friction force exerted on the block?
b. How much will the friction force be if a horizontal force 5N is exerted on the block?
c. What is the minimum force that will start the block in motion?
d. What is the minimum force that will keep the block in motion once it has been started?
e. If the horizontal force is 10N, what is the friction force?

a. $$F_{fric} = \mu_s W = 0.4*20 = 8N $$

b. Since $$F_{applied} < F_{fric}$$

$$F_{fric} = 5N$$

c. Minimum force required to start the motion $$F = F_{static fric} = 8N$$

d. Minimum force required to start the motion

$$F = F_{kineticfric} = \mu_kW = 0.2*20 = 4N$$

e. $$F > F_{staticfric}$$ motion started.

$$F > F_{kineticfric}$$ hence motion continues.

$$F_{fric} = 4N$$

An object of mass 10 kg is to be kept at rest on an inclined plane making an angle of $$37^0$$ to the horizontal by applying a force F along the plane upwards as shown in Fig. The coefficient of static friction between the object and the plane is 0.Find the magnitude of force F.
[Take g= 10 $$ms^{-2}$$]

$$mg sin 37^0= 10\times 10\times 3/5 = 60 N$$
$$f_l=\mu = 0.2 mg cos 37^0$$
$$=0.2 \times 10 \times 10\times 4/5$$
=16N
Friction force alone is unable to balance $$mg sin37^0$$. Hence, F is required for balancing. Here two cases arise.
Case I:
If the block has tendency to slide down, friction on block will act upwards.
F + 16 = 60
$$Rightarrow F = 44N$$
This is the minimum force required for balancing.
Case II: If the block has on tendency to slide up, friction on block will act downwards.
F = 60 + 16$$\Rightarrow$$ F= 76 N
This is the maximum force required for balancing.
Therefore, for balancing $$44N \leq F \leq 16 N$$
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A block of mass m= 2 kg is kept on a rough horizontal surface. A horizontal force F= 5N $$\mu$$ = 0.25 is just able to slide the block. Find the coefficient of static friction. If F= 4N, then what is the frictional force acting on the block?

R.E.F image

$$ \mu = 0.25 $$

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A block of mass $$2 \,kg$$ is pushed against a rough vertical wall with a force of $$40 \,N$$, coefficient of static friction being $$0.5$$. Another horizontal force of $$15 \,N$$, is applied on the block in a direction parallel to the wall. Will the frictional force exerted by the wall on the block ?

$$N_1 = 40N$$

$$F_5 = \mu N_1$$

$$= 0.5 \times 40 = 20 \,N$$

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A uniform chain is just at rest over a rough horizontal table with its $$\frac{I}{\eta}th$$ part of length hanging vertically. Find the of co-efficient of static friction between the chain and the table.

R.E.F image

Total Length L

Linear mass density $$ \lambda = \dfrac{M}{L} $$

mass of hanging portion $$ = \lambda \dfrac{L}{n} $$

$$ = \dfrac{M}{n} $$

Mass of charh on table $$ = \lambda (L-\dfrac{L}{n}) $$

$$ = m\dfrac{(n-1)}{n} $$

As it just rest over table $$ f = $$ weight of hanging portion

$$ \mu (N) = \dfrac{M}{n}g $$

$$ \mu \,M\dfrac{(n-1)}{n}g = \dfrac{M}{n}g $$

$$\boxed{ \mu = \dfrac{1}{n-1} }$$ Ans

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A car and a lorry are travelling with same velocity on a straight horizontal road. Which of the two has got greater momentum?

Lorry, as the mass of lorry is greater then the mass of car.

A force of $$98N$$ is just able to move a block of mass $$20kg$$ on a rough horizontal surface. Calculate the coefficient of friction and the angle of friction ($$\gamma=9.8{ms}^{-2}$$)

Here , maximum possible value of friction = $$\mu mg$$

= $$98$$ $$N$$

We get , coefficient of friction , $$\mu$$ = $$0.5$$.

Now , angle of friction = $$tan^{-1}(\mu )$$

= $$tan^{-1}(0.5 )$$

= $$26.56^{\circ}$$

A mass of $$10kg$$ just pulled along a rough horizontal surface when a force of $$58.8N$$ is applied parallel to the surface. Find the coefficient of static friction between the surfaces

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A blog of mass m is placed on a cart of mass M as shown in the figure, The coefficient of friction between the block and the cart is $$\mu$$. What is the minimum acceleration of the cart so that block m does not slip on the cart?

Given, Mass of the block is $$m$$, MAss of the cart is $$M$$, $$Coefficient=\mu$$

If the cart $$f$$ accelerate with 'a' Then, the pseudo force on the mass $$F=ma$$

Force of friction is $$\mu N$$

For horizontal equilibrium $$F=N=\mu ma$$

The block will not fall as,

$$f\geq mg\Rightarrow \mu ma \geq mg$$

$$\Rightarrow a\geq \dfrac{g}{\mu}$$

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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 $$10\ m s^{-2}$$.

Given : $$m = 10 \ kg$$ $$u = 80 \ cm = 0.8 \ m$$

Velocity before it strikes the ground $$v = \sqrt{2gh}$$

$$\implies \ v = \sqrt{2\times 10\times 0.8} = 4 \ m/s$$

Momentum transfer $$P = mv = 10\times 4 = 40 \ kg \ m/s$$

While catching a fast moving cricket ball a fielder in the ground gradually pulls his hand backward with the ball. Explain.

While catching a fast moving cricket ball, a fielder in the ground gradually pulls his hand backward because by doing this, he increases the time in which the velocity of ball will become zero. Thus, the rate of change of momentum will be large and less force will be exerted by the ball on the hand of the fielder. Hence, he will not get injured by the ball.

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Light in certain cases may be considered as a stream of particles called photons. Each photon has a linear momentum $$\cfrac{h}{\lambda }.$$ where $$h$$ is the Planck's constant and $$\lambda $$ is the wavelength of the light.A beam of light of wavelength $$\lambda $$ is incident on a plane mirror at an angle of incidence $$\theta $$. Calculate the change in linear momentum of photon as the beam is reflected by the mirror.

Initial momentum of photon =$$ \cfrac{h}{\lambda} \cos \theta \hat i - \cfrac{h}{\lambda} \sin \theta \hat j$$

Final momentum of photon=$$ \cfrac{-h}{\lambda} \cos \theta \hat i - \cfrac{h}{\lambda} \sin \theta \hat j$$

$$ \therefore $$ Change in linear momentum = Final - Initial

$$ = \cfrac{-2h}{\lambda} \cos \theta \hat i$$

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A plank with a box on it at one end is gradually raised about the other end.As the angle of inclination with the horizontal reaches $$30^o$$ , the box starts to slip and slides $$4m$$ down the plank in $$4s$$. The coefficient of static and kinetic friction between the box and the plank will be respectively

At $$ \theta = 30^o$$

Box is about to slip, hence

$$ m g \sin \theta = \mu _s mg \cos \theta$$

$$ \therefore \mu_s = \tan \theta$$

$$ = \tan 30^o = \cfrac{1}{\sqrt3}$$

Now, $$ s= ut + \cfrac{1}{2} at^2$$

$$ 4 = 0 + \cfrac{1}{2} a \times 4^2$$

$$ \therefore a = \cfrac{8}{16} = \cfrac{1}{2}$$

But $$ a = \cfrac{ mg\sin \theta - \mu_k mg \cos \theta}{m}$$

$$ \cfrac{1}{2} = g[\sin 30^o - \mu_k \cos 30^o]$$

$$ \cfrac{1}{20} = \cfrac{1}{2} - \mu_k \times \cfrac{\sqrt 3}{2}$$

$$ \therefore \mu_k \times \cfrac{\sqrt 3}{2} = \cfrac{1}{2} - \cfrac{1}{20}$$

$$ \therefore \mu_k = \cfrac{9}{20} \times \cfrac{2}{\sqrt 3}$$

$$ = \cfrac{3 \sqrt 3}{10}$$

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Coefficient of static friction depends on ____________

The coefficient of static frictional force depends upon the nature of the surfaces in contact and on their roughness. It is independent of the size or area of the surfaces

A ball of mass $$200$$ g is projected with a velocity of 30$$\mathrm { m } / \mathrm { s }$$ at $$30 ^ { \circ }$$ from horizontal. Using the concept of impulse, find change in velocity in $$2$$ s. Take $$g = 10 m / s ^ { 2 }$$

Impulse will be provided by gravity and in the direction of this force that is $$downward.$$

impulse will be $$\Delta P$$ given by $$\Delta P=F\Delta t=mg\times 2sec=0.2kg\times 10m/s^2\times 2sec=4N-s$$

This impulse is nothing but change in $$momentum.$$

so change in velocity will be $$\Delta v=\dfrac{\Delta P}{m}=\dfrac{4}{0.2}=20m/s$$ in $$downward $$ direction.

Two masses A and B of 10 Kg and 5 Kg respectively are connected with a string passing over a frictionless pulley at a corner of a table a shown in the adjoining diagram. The coefficient of friction of A with the table is 0.The minimum mass of C that may be placed on A to prevent it from moving is equal to?

Given that,

Mass of A $${{m}_{A}}=10\,kg$$

Mass of B $${{m}_{B}}=5\,kg$$

Coefficient of friction $$\mu =0.2$$

Now, the tension is

$$ T={{m}_{b}}g $$

$$ T=5\times 10 $$

$$ T=50\,N $$

Now, we consider the combination of A and C

Then,

$$ T=\mu R $$

$$ T=0.2\left( {{m}_{A}}+{{m}_{C}} \right)g $$

$$ 50=0.2\times 10\times 10+0.2{{m}_{C}}\times 10 $$

$$ 50-20=2{{m}_{c}} $$

$$ {{m}_{c}}=15\,kg $$

Hence, the minimum mass of $$C$$ is $$15\ kg$$

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The upper portion of an inclined plane of inclination a is smooth and the lower portion is rough. A particle slide down from rest from the top and just comes to rest at the foot. If the ratio of the smooth length to rough length is m:n, the coefficient of friction is

the work was done by all the forces acting on the particle, and equating the total to zero.

Assume the total length of the plane is $$ (m+n)d$$, the coefficient of friction is $$μ$$ and the mass of the particle is .$$M$$.

The acceleration of the particle at right angles to the plane is zero. So, resolving in this direction, the normal contact force N between the particle and the plane is given by

$$N=mgcosα$$

and, since the particle moves at right angles to this force, it does a zero amount of work.

During the second stage the friction force acting on the particle is $$μN= μmgcosα.$$ In this phase, the particle moves through a distance and down the plane, and so the work done on the particle by this friction force is, therefore -.μmgndcosα

The particle falls through a total height (m+n)dsinα. The work done on the particle by its weight is, therefore, $$Mg(m+n)dsinα$$

Since the particle starts and finishes at rest, the gain in its KE is zero, and therefore the total amount of work done on the particle is zero. So we have:

$$-μmg cosα+Mg(m+n)dsinα=0$$

$$μ=(m+n)tanα/n$$

The position (x) - time (t) graph for a particle of mass $$1 kg$$ moving along x-axis is shown in figure. Find impulse on particle at $$t = 4 s$$.

Velocity is nothing but the $$slope $$ of $$x-t$$ graph, so the velocity just before $$t=4$$ will be

$$V_1=slope=\dfrac{(20-0)m}{(4-0)sec}=5m/sec$$

the velocity just after $$t=4$$ will be

$$V_2=slope=\dfrac{(0-20)m}{(8-4)sec}=-5m/sec$$

so the change in momentum at $$t=4sec$$ will be $$\Delta P=m(V_2-V_1)=1kg(-5-5)m/s=-10Kgm/s$$

if considering only magnitude the impulse is $$10Kgm/s$$ or $$10N-sec$$

Calculate the change in momentum of a body weighing 5 kg when its velocity decreases from 20 m/s to 0.20 m/s

Mass of body=5 kg
Velocity $$v_1=20 m/s$$
Momentum $$p_1=20\times 5=100 kg.m/s$$
Velocity $$v_2=0.2 m/s$$
Momentum $$p_2=5 \times 0.2=1 kg.m/s$$
Change in momentum =$$p_2-p_1=1-100=-99 kg.m/s$$(Negative sign shows that momentum decreases)

Define momentum of a body. On what factor does the momentum of a body depend?

Momentum is the physical quantity which is the measure of the quantity of motion of a moving body. It depends on mass and velocity of the body.

What will be the momentum of a body of mass 'm' which is moving with a velocity?

Momentum of a body of mass 'm' and velocity 'v' will be $$p=m \times v$$

Differentiate between Motion and force of friction.

Motion	Friction
1. Motion is a change in position or displacement of an object with respect to time. Motion is mathematically described in terms of displacement, distance, velocity, acceleration, time, and speed.	1. Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other.
2. Contact between the objects is not required for motion of the object.	2. Contact between the objects is required so that friction can come into existence.
3. For example: pushing a block leads to its motion in the direction of applied force, throwing a ball etc.	3.For example: walking on the ground , riding a bicycle etc.

Why are car seat-belts designed to stretch somewhat in a collision?

Car seat-belts are somewhat stretchable so as to increase the time taken by the passengers to fall forward.Due to this,the rate of change of momentum of passengers is reduced and hence less stopping force acts on them.So the passengers do not get hurt.

A block of mass $$15\ kg$$ is resting on a rough inclined plane as shown in figure. The block is tied up by a horizontal string having a tension of $$50\ N$$. Calculate the coefficient of friction between the block and the inclined plane.

$$M=15kg$$

$$T=50N$$

Equating forces

$$f=\mu[T \sin \theta+mg\cos\theta]$$

$$mg\sin\theta=T\cos\theta+f$$

$$f=\mu[50/\sqrt{2}+150/\sqrt{2}]$$

$$150/\sqrt{2}=50/\sqrt{2}+f$$

$$\dfrac{100}{\sqrt{2}}=f$$

$$\mu[\dfrac{200}{\sqrt{2}}]=\dfrac{100}{\sqrt{2}}$$

$$\mu=0.5$$

A child places a picnic basket on the outer rim of a merry-go-round that has a radius of $$4.6$$ m and revolves once every $$30$$ s.
(a) What is the speed of a point on that rim? (b) What is the lowest
value of the coefficient of static friction between the basket and
merry-go-round that allows the basket to stay on the ride?

(a) The distance traveled in one revolution is $$2\pi R = 2\pi(4.6 m) = 29$$ m. The (constant)
speed is consequently $$v = (29 m)/(30 s) = 0.96\, m/s$$.
(b) Newton’s second law ( for the magnitude of the acceleration) leads to
$$f_s=m(\dfrac{v^2}{R})=m(0.20)$$
in SI units. Noting that $$F_N = mg$$ in this situation, the maximum possible static friction is $$f_{s,max} = μ_s mg$$. Equating this with $$f_s = m(0.20)$$ we find the mass m cancels
and we obtain $$μ_s = 0.20/9.8 = 0.021$$.

A cat dozes on a stationary merry-go-round in an amusement park, at a radius of $$5.4 m$$ from the center of the ride. Then the operator turns on the ride and brings it up to its proper turning rate of one complete rotation every $$6.0$$ s. What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding (or the cat clinging with its claws)?

Perhaps surprisingly, the equations pertaining to this situation are exactly those in

Sample Problem – “Car in flat circular turn,” although the logic is a little different. In the
Sample Problem, the car moves along a (stationary) road, whereas in this problem the cat
is stationary relative to the merry-go-round platform. But the static friction plays the
same role in both cases since the bottom-most point of the car tire is instantaneously at
rest with respect to the race track, just as static friction applies to the contact surface
between cat and platform. we find,
$$\mu_s(2\pi R/T)^2/gR=4\pi^2R/gT^2 $$
With $$T = 6.0$$ s and $$R = 5.4$$ m, we obtain $$\mu_s = 0.60$$

Answer the following in short.
When you push something heavy, the resistance you feel decreases as soon as the object starts moving. Explain the reason.

It is because when the object is at rest, static friction occurs. The static friction is maximum when the applied force is just short of the force required to start motion between the two bodies in contact. Once the object starts moving, friction (resistance) we feel decreases.

Name the three kinds of friction.

Friction is of three kinds:
1. Static friction
2. Sliding friction,
3. Rolling friction.

When the cutting edge of a knife is put against a fast
rotating stone to sharpen it, sparks are seen to fly. Explain the
reason.

When a knife is sharpened by putting it against a rotating stone, it produces heat. With the increase in the speed of the rotating stone heat on the knife also increases which makes it spark.

A body X of mass $$5kg$$ is moving with velocity $$20ms^{1}$$ while another body Y of mass $$20kg$$ is moving with velocity $$5ms^{1}$$. Compare the momentum of the two bodies.

For body X,

mass:$$m_X=5\ kg$$

velocity: $$v_x= 20\ m/s$$

$$\therefore$$momentum:$$p_X=m_X \times v_X=5 \times 20 = 100\ kg\ m/s$$

For body Y,

mass:$$m_Y=20\ kg$$

velocity: $$v_Y= 5\ m/s$$

$$\therefore$$momentum: $$p_Y=m_Y \times v_Y=20 \times 5 = 100\ kg\ m/s$$

So,$$p_X:p_Y=1:1$$

Define the term linear momentum. State its S.I. unit.

The product of mass and velocity of the body is called linear momentum.

S.I. unit of linear momentum is kg m $$s^{-1}$$.

Write an expression for the change in momentum of a body of mass m moving with velocity v if (i) $$v << c$$ and (ii) $$v \rightarrow c$$.

The product of mass and velocity of the body is called linear momentum.

$$P=mv$$

(i)When $$v<c,$$ $$\Delta$$p=m$$\Delta$$v

(ii) When $$v \rightarrow c$$, the change in momentum is $$\Delta p=\Delta(mv)$$

Name the S.I. unit of rate of change in momentum.

Rate of change in moemtum = momentum/time

$$=\dfrac{P}{t}=\dfrac{mv}{t}=ma=F$$

so Rate of change in moemtum is force and it's unit is N.

rate of change in momentum is newton(N).

State the condition when the change in the momentum of a body depends only on the change in its velocity.

The product of mass and velocity of the body is called linear momentum.

If velocity v of the moving body is much smaller than the velocity of light c (v<c) then momentum doesn't depend on mass, it depends on velocity only

A rocket is moving at a constant speed in space by burning its fuel and ejecting out the burnt gases through a nozzle. Answer the following:
Is there any change in the momentum of the rocket? If yes. what causes the change in momentum?

Momentum = mass x veocity

$$P=mv$$

Yes. there is a change in momentum of the rocket. The decrease in mass of

the rocket causes a change in momentum.

$$\Delta P = \Delta m v$$

A car of mass $$600$$ kg is moving with a speed of $$10\, ms^{-1}$$ while a scooter of mass $$80kg$$ is moving with a speed of $$50 ms^{-1}$$.Compare their momentum.

Mass of the car, $$M = 600\,kg$$

Mass of the scooter,$$m = 80\,kg$$

Velocity of car. $$v = 10 m/s$$

Velocity of Scooter $$= 50 m/s$$

we know that $$Momentum = mass \times Velocity$$

$$\dfrac{\text{momentum of car}}{\text{momentum of scooter}}=\dfrac{MV}{mv}=\dfrac{(600)(10)}{(80)(50)}=\dfrac{3}{2}$$

Thus. the required ratio is $$3:2$$

A.B and C are the three forces each of magnitude $$4$$ n acting in the plane of the paper as shown in the figure. The point O lies in the same plane.
which force has the greatest moment about O? Give a reason.

We know that,

Moment of force = Force $$\times$$ Perpendicular distance

Since the perpendicular distance of vector A is greatest from point O,

Vector A will have the greatest moment about O.

A force of $$10$$ kgf is applied on a body of mass $$100$$ g initially at rest for $$0.18$$ calculate the momentum acquired by the body.

Mass of body$$= 100 g=0.1 \ kg$$

Initial velocity.$$u=0 \ m/s$$

Time ,$$\Delta t = 0.1s$$

Force $$\times$$ change in time = change in momentum

Force.$$F = 10 \ kgf = 10 \times 10=100N$$

($$\therefore 1kgf=10 \ N$$)

$$F \times \Delta t = \Delta p$$ (change in momentum)

$$100 \times 0.1 = \Delta p$$

$$\Delta p=10\ kg \ m/s$$

Define momentum. Prove that momentum before the collision is equal to the total momentum after the collision.

Momentum: The product of mass and velocity of a moving body is called momentum.
Momentum = Mass of body $$\times$$ Velocity.
$$\vec{p} = m \vec{v}$$
Law of conservation of momentum: The law states, in absence of external forces the momentum of the system is conserved. It means that the total momentum before the collision is equal to the total momentum after the collision. Suppose two bodies A and B of masses $$m_1$$ and $$m_2$$ are moving in a straight line in the same direction with velocities $$u_1$$ and $$u_2$$, the velocity of the body A is more than that of B. So both the bodies will collide after some time. Velocities of the bodies A and B after collision become $$v_1$$ and $$v_2$$, respectively. The time of collision is t.
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Define coefficient of static friction.

The coefficient of static friction is defined as the ratio of limiting friction to the normal reaction between the surfaces.

Distinguish between static friction, limiting friction & kinetic friction. Now do they vary with applied force? Explain.

The static friction is a friction that acts on a body at rest.
Limiting friction is the maximum value of static friction. It is the force that is required for the body to just start moving.
Kinetic friction is the frictional force that action a body which is in motion.
On increasing the applied force, static friction increase, until it reaches limiting friction which is fixed, and kinetic friction also remains constant.
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A loaded lorry of mass $$12000\ kg$$ moves with a velocity of $$12\ m/s$$. Its velocity becomes $$10\ m/s$$ after $$5\ s$$
What is the change in momentum?

change momentum $$=mv-mu$$

$$(12000\times10)-(12000\times 12)$$

$$=120000-144000$$

$$=-24000\ kg\ m/s$$

A hockey ball of mass $$200\ g$$ hits on a hockey stick with a velocity $$10\ m/s$$. Calculate the change in momentum if the ball bounces back on the same path with the same speed.

$$m=200g=200/1000=0.2\ kg$$
initial momentum $$=mu$$
$$=0.2 \times 10=2\ kg\ m/s$$
Final momentum $$=mv$$
$$=0.2 \times 70=2\ kg m/s$$
Change in momentum $$=mv-mv=-2-2$$
$$=-4\ kg\ m/s$$

A loaded lorry of mass $$12000\ kg$$ moves with a velocity of $$12\ m/s$$. Its velocity becomes $$10\ m/s$$ after $$5\ s$$
What is the initial momentum and what is the final momentum?

initial momentum $$=mu$$
$$=12000 \times 12=144000\ kg\ m/s$$

Final momentum $$=mv$$
$$=12000 \times 10=120000\ kg\ m/s$$

A loaded lorry of mass $$12000\ kg$$ moves with a velocity of $$12\ m/s$$. Its velocity becomes $$10\ m/s$$ after $$5\ s$$
What is the rate of change of momentum?

Rate the change in momentum

$$\Delta p=\dfrac{mv-mu}{t}$$

$$\Delta p=\dfrac{(12000\times10)-(12000\times12)}{5}$$

$$=\dfrac{-24000}{5}$$

$$=-4800\ kg m/s^2$$

What is meant by limiting friction.

Limiting friction is the force of static friction acting just before the body starts moving.

A $$15.0-m$$ uniform ladder weighting $$500\ N$$ rests against a frictionless wall. The ladder makes a $$60^o$$ angle with the horizontal.
If the ladder is just on the verge of slipping when the firefighter is $$9.00\ m$$ from the bottom, what is the coefficient of static friction between ladder and ground?

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A uniform chain of mass $$M$$ and Length $$L$$ is held vertically in such a way that its lower end just touches the horizontal floor. The chain is released from rest in this position. Any portion that strikes the floor comes to rest. Assuming that the chain does not form a heap on the floor, calculate the force exerted by it on the floor, when a length $$x$$ has reached the floor.

As chain has slip $$x$$ from both sides then total $$2x$$ length will be under the table.

mass under the table=$$\dfrac{m}{l}\times 2x$$

So, acceleration, $$=\dfrac{\dfrac{2mx}{L}g}{M}$$

$$a=\dfrac{2gx}{l}$$

Block $$M$$ slides down a frictionless incline as shown. Find the minimum friction coefficient so that $$m$$ does not slide with respect to $$M$$.

Let the slope of the inclination is $$\theta$$.

Hence component of weight along the slide $$= mg\sin{\theta}$$

Normal component of weight , $$N = mgcos{\theta}$$

$$\Rightarrow $$ force of friction $$= F = \mu N = \mu mg\cos{\theta}$$

For the block does not slide $$F = mg\sin{\theta}$$

$$\Rightarrow \mu mg\cos{\theta} = mg\sin{\theta}$$

$$\Rightarrow \mu \cos{\theta}=\sin{\theta}$$

$$\Rightarrow \mu = \tan{\theta}$$

So the block does not slide if the co-efficient of the friction will be $$\tan{\theta}$$

A uniform chain of length $$\ell$$ is placed on a rough table, with length $$\ell/3$$ hanging over the edge. If the chain just begins to slide off the table by itself from this portion, the coefficient of friction between the chain and the table is $$1/K$$. Then, find the value of $$K$$.

We know,

For this case to happen,

$$F_g=f_{max}$$

Let Linear mass density be $$\lambda$$

$$\lambda\dfrac{l}{3}g=\mu\lambda\dfrac{2l}{3}g$$

$$\dfrac{1}{3}=\dfrac{1}{K}\dfrac{2}{3}$$

$$K=2$$

What is momentum? Write its SI unit. Interpret force in terms of momentum. Represent the following graphically.

Momentum The quantity of motion possessed by a moving body is known as momentum of the body.

Momentum of an object of mass m moving with a velocity v is the product of it.
It is denoted by p

p = mv
Its SI unit is kg-ms^-1.
As from Newton’s llnd law, rate of change of momentum is equal to force applied.

Two forces of $$6N$$ and $$8N$$ act at a point at an angle of $${90^ \circ }$$ with each other. Determine the magnitude and direction of the resultant force by drawing a vector diagram.

2045901_1347493_ans_fd7bea137c3b44dca625186279ac193b.jpg

A machine gun has a mall of 20 kg. If fires 35 g bullets at the vale of 4 bullets $${ P }^{ n }$$ second, with a spend of $$400 m{ s }^{ -1 }$$. What force must be applied to the gun to help it in position ?

2046082_1388201_ans_94ddad78c8b140f9832488df6e8ecef2.jpg

A coin placed $$30.0 cm$$ from the center of a rotating, horizontal turntable slips when its speed is $$50.0 cm/s$$. (a) What force causes the centripetal acceleration when the coin is stationary relative to the turntable? (b) What is the coefficient of static friction between coin and turntable?

Below figure shows the constant maximum speed of the turntable and the centripetal acceleration of the coin.
(a) The force of static friction causes the centripetal acceleration.
(b) From below figure,
$$ma\hat{i}=f\hat{i}+n\hat{j}+mg(-\hat{j})$$
$$\sum F_{y}=0-n-mg$$
thus $$n=mg$$ and
$$\sum F_{r}=m\dfrac{v^{2}}{r}=f=\mu n=\mu mg$$
Then,
$$\mu=\dfrac{v^{2}}{rg}=\dfrac{(50.0cm/s)^{2}}{(30.0cm)(980cm/s^{2})}=0.0850$$
1801933_1860176_ans_e8aacce68f144aef805c3425a4292fe9.png

1801933_1860176_ans_e8aacce68f144aef805c3425a4292fe9.png

The mayor of a city reprimands some city employees because they will not remove the obvious sags from the cables that support the city traffic lights. What explanation can the employees give? How do you think the case will be settled in mediation.

It is impossible to string a horizontal cable without its sagging a bit. Since the cable has a mass, gravity pulls it downward. A vertical component of the tension must balance the weight for the cable to be in equilibrium. If the cable were completely horizontal, then there would be no vertical component of the tension to balance the weight. If a physicist would testify in court, the city employees would win.

A cannon is rigidly attached to a carriage,which can move along horizontal rails but is connected to a post by a large spring, initially unstretched and with force constant $$k=2.00 \times 10^4 N/m$$, as shown in Figure $$P9.70$$. The cannon fires a $$200-kg$$ projectile at a velocity of $$125\ m/s$$ directed $$45.0^o$$ above the horizontal.
Assuming that the mass of the cannon and its carriage is $$5\ 000\ kg$$. find the recoil speed of the cannon.

Given,

spring constant $$k=2\times { 10 }^{ 4 }N/m$$

velocity of ball is $$V=125m/s$$

mass of ball is $$m=200kg$$

angle of projection is $${ 45 }^{ \circ }$$

mass of cannon and its carriage is $$5000kg$$

Here, momentum is conserve because gravity and normal force are acting in $$x$$ direction therefore, we can write,

$${ P }_{ i }={ P }_{ f }\\ m{ v }_{ i }cos45=(m+M){ v }_{ f }$$

Substitute all value in above equation

$$200\times 125\times \frac { 1 }{ \sqrt { 2 } } =5000{ v }_{ f }\\ { v }_{ f }=\frac { 200\times 125 }{ 5000\times \sqrt { 2 } } \\ { v }_{ f }=3.54m/s$$

Hence, recoil speed of the cannon is $$3.54m/s$$

Laws Of Motion - Class 11 Medical Physics - Extra Questions

Coefficient of static friction is depends on _______________

Fill in the blanks:Friction is a type of ________.

Find the moment of force of $$ 20 N $$ about an axis of rotation at distance $$ 0.5 $$ m from the force.

Which are the following occation friction is beneficial and non beneficialstriking a match sticks on the side of a match boxwear and tear of machines.ability to hold objectswearing out of tyreswalking $$6.$$ loss of fuel

Acceleration of block A and B.

A hockey ball of mass $$200$$g travelling at $$10$$m $$s^{-1}$$ is struck by a hockey stick so as to return it along its original path with a velocity at $$5$$m $$s^{-1}$$. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

A block of mass m= 2 kg is kept on a rough horizontal surface. A horizontal force F= 5N $$\mu$$ = 0.25 is just able to slide the block. Find the coefficient of static friction. If F= 4N, then what is the frictional force acting on the block?

A uniform chain is just at rest over a rough horizontal table with its $$\frac{I}{\eta}th$$ part of length hanging vertically. Find the of co-efficient of static friction between the chain and the table.

A car and a lorry are travelling with same velocity on a straight horizontal road. Which of the two has got greater momentum?

A force of $$98N$$ is just able to move a block of mass $$20kg$$ on a rough horizontal surface. Calculate the coefficient of friction and the angle of friction ($$\gamma=9.8{ms}^{-2}$$)

A mass of $$10kg$$ just pulled along a rough horizontal surface when a force of $$58.8N$$ is applied parallel to the surface. Find the coefficient of static friction between the surfaces

A blog of mass m is placed on a cart of mass M as shown in the figure, The coefficient of friction between the block and the cart is $$\mu$$. What is the minimum acceleration of the cart so that block m does not slip on the cart?

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 $$10\ m s^{-2}$$.

While catching a fast moving cricket ball a fielder in the ground gradually pulls his hand backward with the ball. Explain.

Coefficient of static friction depends on ____________

A ball of mass $$200$$ g is projected with a velocity of 30$$\mathrm { m } / \mathrm { s }$$ at $$30 ^ { \circ }$$ from horizontal. Using the concept of impulse, find change in velocity in $$2$$ s. Take $$g = 10 m / s ^ { 2 }$$

The upper portion of an inclined plane of inclination a is smooth and the lower portion is rough. A particle slide down from rest from the top and just comes to rest at the foot. If the ratio of the smooth length to rough length is m:n, the coefficient of friction is

The position (x) - time (t) graph for a particle of mass $$1 kg$$ moving along x-axis is shown in figure. Find impulse on particle at $$t = 4 s$$.

Calculate the change in momentum of a body weighing 5 kg when its velocity decreases from 20 m/s to 0.20 m/s

Define momentum of a body. On what factor does the momentum of a body depend?

What will be the momentum of a body of mass 'm' which is moving with a velocity?

Differentiate between Motion and force of friction.

Why are car seat-belts designed to stretch somewhat in a collision?

A block of mass $$15\ kg$$ is resting on a rough inclined plane as shown in figure. The block is tied up by a horizontal string having a tension of $$50\ N$$. Calculate the coefficient of friction between the block and the inclined plane.

Answer the following in short.When you push something heavy, the resistance you feel decreases as soon as the object starts moving. Explain the reason.

Name the three kinds of friction.

When the cutting edge of a knife is put against a fastrotating stone to sharpen it, sparks are seen to fly. Explain thereason.

A body X of mass $$5kg$$ is moving with velocity $$20ms^{1}$$ while another body Y of mass $$20kg$$ is moving with velocity $$5ms^{1}$$. Compare the momentum of the two bodies.

Define the term linear momentum. State its S.I. unit.

Write an expression for the change in momentum of a body of mass m moving with velocity v if (i) $$v << c$$ and (ii) $$v \rightarrow c$$.

Name the S.I. unit of rate of change in momentum.

State the condition when the change in the momentum of a body depends only on the change in its velocity.

A rocket is moving at a constant speed in space by burning its fuel and ejecting out the burnt gases through a nozzle. Answer the following:Is there any change in the momentum of the rocket? If yes. what causes the change in momentum?

A car of mass $$600$$ kg is moving with a speed of $$10\, ms^{-1}$$ while a scooter of mass $$80kg$$ is moving with a speed of $$50 ms^{-1}$$.Compare their momentum.

A.B and C are the three forces each of magnitude $$4$$ n acting in the plane of the paper as shown in the figure. The point O lies in the same plane.which force has the greatest moment about O? Give a reason.

A force of $$10$$ kgf is applied on a body of mass $$100$$ g initially at rest for $$0.18$$ calculate the momentum acquired by the body.

Define momentum. Prove that momentum before the collision is equal to the total momentum after the collision.

Define coefficient of static friction.

Distinguish between static friction, limiting friction & kinetic friction. Now do they vary with applied force? Explain.

A loaded lorry of mass $$12000\ kg$$ moves with a velocity of $$12\ m/s$$. Its velocity becomes $$10\ m/s$$ after $$5\ s$$What is the change in momentum?

A hockey ball of mass $$200\ g$$ hits on a hockey stick with a velocity $$10\ m/s$$. Calculate the change in momentum if the ball bounces back on the same path with the same speed.

A loaded lorry of mass $$12000\ kg$$ moves with a velocity of $$12\ m/s$$. Its velocity becomes $$10\ m/s$$ after $$5\ s$$What is the initial momentum and what is the final momentum?

A loaded lorry of mass $$12000\ kg$$ moves with a velocity of $$12\ m/s$$. Its velocity becomes $$10\ m/s$$ after $$5\ s$$What is the rate of change of momentum?

What is meant by limiting friction.

Block $$M$$ slides down a frictionless incline as shown. Find the minimum friction coefficient so that $$m$$ does not slide with respect to $$M$$.

A uniform chain of length $$\ell$$ is placed on a rough table, with length $$\ell/3$$ hanging over the edge. If the chain just begins to slide off the table by itself from this portion, the coefficient of friction between the chain and the table is $$1/K$$. Then, find the value of $$K$$.

What is momentum? Write its SI unit. Interpret force in terms of momentum. Represent the following graphically.

Two forces of $$6N$$ and $$8N$$ act at a point at an angle of $${90^ \circ }$$ with each other. Determine the magnitude and direction of the resultant force by drawing a vector diagram.

A machine gun has a mall of 20 kg. If fires 35 g bullets at the vale of 4 bullets $${ P }^{ n }$$ second, with a spend of $$400 m{ s }^{ -1 }$$. What force must be applied to the gun to help it in position ?

A coin placed $$30.0 cm$$ from the center of a rotating, horizontal turntable slips when its speed is $$50.0 cm/s$$. (a) What force causes the centripetal acceleration when the coin is stationary relative to the turntable? (b) What is the coefficient of static friction between coin and turntable?

The mayor of a city reprimands some city employees because they will not remove the obvious sags from the cables that support the city traffic lights. What explanation can the employees give? How do you think the case will be settled in mediation.

Class 11 Medical Physics Extra Questions

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Fill in the blanks:
Friction is a type of ________.

Which are the following occation friction is beneficial and non beneficial
striking a match sticks on the side of a match box
wear and tear of machines.
ability to hold objects
wearing out of tyres
walking $$6.$$ loss of fuel

Answer the following in short.
When you push something heavy, the resistance you feel decreases as soon as the object starts moving. Explain the reason.

When the cutting edge of a knife is put against a fast
rotating stone to sharpen it, sparks are seen to fly. Explain the
reason.

A rocket is moving at a constant speed in space by burning its fuel and ejecting out the burnt gases through a nozzle. Answer the following:
Is there any change in the momentum of the rocket? If yes. what causes the change in momentum?

A.B and C are the three forces each of magnitude $$4$$ n acting in the plane of the paper as shown in the figure. The point O lies in the same plane.
which force has the greatest moment about O? Give a reason.

A loaded lorry of mass $$12000\ kg$$ moves with a velocity of $$12\ m/s$$. Its velocity becomes $$10\ m/s$$ after $$5\ s$$
What is the change in momentum?

A loaded lorry of mass $$12000\ kg$$ moves with a velocity of $$12\ m/s$$. Its velocity becomes $$10\ m/s$$ after $$5\ s$$
What is the initial momentum and what is the final momentum?

A loaded lorry of mass $$12000\ kg$$ moves with a velocity of $$12\ m/s$$. Its velocity becomes $$10\ m/s$$ after $$5\ s$$
What is the rate of change of momentum?