A \\(5000\ kg\\) rocket is set for vertical firing. The exhaust speed is \\(800\ ms^{-1}.\\) To give an upward acceleration of \\(20\ ms^{-2},\\) the amount of gas ejected per second to supply the needed thrust is \\(\left ( g = 10\ ms^{-2} \right )\\)

Given that, Mass of rocket \\(m=5000\,Kg\\) Acceleration \\(a=20\,m/{{s}^{2}}\\) Speed \\(v=800\,m/s\\) \\(g=10\,m/{{s}^{2}}\\) Now, thrust force on the rocket \\({{F}_{t}}={{v}_{r}}\left( \dfrac{-dm}{dt} \right)\\) Net force on the rocket \\( {{F}_{net}}={{F}_{t}}-W \\) \\( ma={{v}_{r}}\left( \dfrac{-dm}{dt} \right)-mg \\) \\( \left( \dfrac{-dm}{dt} \right)=\dfrac{m\left( g+a \right)}{{{v}_{r}}} \\) Rate of gas ejected per second \\( \left( \dfrac{-dm}{dt} \right)=\dfrac{5000\left( 10+20 \right)}{800} \\) \\( \dfrac{-dm}{dt}=187.5\,kg/s \\) Hence, the amount of gas ejected per second is \\(187.5\ kg/s\\)

A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 m/s. When the stone reaches the floor,the distance of the man above the floor will be

Let the speed of the man in upward direction be \\(V\\).Applying conservation of linear momentum of man and stone system : \\(P_f = P_i\\)\\(M_{man}V - M_{stone}V_{stone} = 0\\)Or \\(50V-0.5\times 2 = 0\\)\\(\implies \ V = 0.02 \ m/s\\)Time taken by stone to reach the ground \\(t = \dfrac{10 \ m}{2 \ m/s} = 5 \ s\\)Distance covered by man in \\(5 \ s\\), \\(d = Vt = 0.02\times 5 = 0.1 \ m\\) (in upward direction)Thus distance of man above the floor \\(h = 10+0.1 = 10.1 \ m\\)

The force 'F' acting on a particle of mass 'm' is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to 8 s is

\\(\Delta P=\left (\dfrac {1}{2}\right )(2)(6)-(3)(2)+(4)(3)\\)\\(=6-6+12\\)\\(=12\\)

A block of metal weighing \\(2\\) kg is resting on a frictionless plane it is struck by a jet releasing water at a rate of \\(1\\) kg/s and at a speed

The final velocity of mass of water after hitting the metal block is zero.So the average change in momentum per sec = initial momentum =mass×initial velocity of water=1kg×5m/s=5kgm/sAs this change in momentum occurs during 1 sec. We may say that the force exerted on the 2kg metal block isF=change in momentumtime=5kgm/s1s=5NHence by Newton's law initial acceleration of the block of mass m=2kg is given bya=Fm=5N2kg=2.5m/s2

A particle moves in the x-y plane under the influence of a force such that its linear momentum is \\( \vec{p}(t)=A(i \cos (k t)-j \sin (k t)] \\) where A and k are constants. The angle between the force andmomentum is

\\(P(t)=A(\hat{j}coskt-\hat{j}sinkt)\\)\\(F=\dfrac{dP(t)}{dt}\\)\\(F=\dfrac{d}{dt}A(-\hat{i}coskt-jsinkt)\\)\\(=A(-\hat{i}sinkt(k)-jcoskt(k))\\)\\(F=Ak(-\hat{i}sinkt-jcoskt)\\)Angle between two vectors is \\(\theta\\)\\(cos\theta=\dfrac{F.P}{|F||P|}\\)\\(=A\dfrac{(\hat{j}cos kt-\hat{j} sin kt).Ak(-\hat{i}sinkt-\hat{j}sinkt)}{|F||P|}\\)\\(\dfrac{A^2k(-sinktcoskt+sinktcoskt)}{|F||P|}\\)\\(=A^2k(0)=0\\)\\(\theta=cos^{-1}(0)=90^0\\)

Two sphere \\(A\\) and \\(B\\) of masses \\(m_{1}\\) and \\(m_{2}\\) respectively collide. \\(A\\) is at rest initially and \\(B\\) is moving with velocity \\(v\\) along x-axis. After collision \\(B\\) has a velocity \\(\dfrac {v}{2}\\) in a direction perpendicular to the original direction. The mass \\(A\\) moves after collision in the direction.

There is no external force acting on the spheres. So linear momentum will be conserved.Before the collision, In direction \\(x\\), Linear momentum \\(= m_2v\\) ....1In direction \\(y\\), Linear momentum\\(=0\\)After the collision, spheres moves as shown in figure. Let velocity of sphere \\(A\\) is \\(v_1\\)In direction \\(x\\), Linear momentum \\(= m_1v_1 cos(\theta)\\) ...2In direction \\(y\\), Linear momentum \\(= \dfrac{m_2v}{2}-m_1v_1sin(\theta)\\) .......3Linear momentum will be conserved, From equation 1 and 2, \\(\Rightarrow m_1v_1cos(\theta)= m_2v\\) ...4From the equation 2, \\(\Rightarrow \dfrac{m_2v}{2}-m_1v_1sin(\theta)=0\\) \\(\Rightarrow m_1v_1 sin(\theta)= \dfrac{m_2v}{2}\\) ......5Dividing equation 5 by 4, \\(\Rightarrow tan(\theta)=\dfrac{1}{2}\\) \\(\Rightarrow \theta= tan^{-1} \left (\dfrac{1}{2}\right ) \\)

State the law of conservation of momentum and derive related expression.

Law of conservation of momentum states that when two objects collide with each other , the sum of their linear momentum always remains same or we can say conserved and is not effected by any action, reaction only in case is no external unbalanced force is applied on the bodies. Let, \\(m_A\\) = Mass of ball A \\(m_B\\)= Mass of ball B \\(u_A\\)= initial velocity of ball A \\(u_B\\)= initial velocity of ball B \\(v_A\\)= Velocity after the collision of ball A \\(v_B\\)= Velocity after the collision of ball B \\(F_{ab}\\)= Force exerted by A on B \\(F_{ba}\\)= Force exerted by B on A Now,Change in the momentum of A= momentum of A after the collision - the momentum of A before the collision = \\(m_A v_A - m_A u_A\\) Rate of change of momentum A= Change in momentum of A/ time taken = \\({\dfrac{m_A v_A - m_A u_A}{t}}\\) Force exerted by B on A (\\(F_{ba}\\)); \\(F_{ba}= {\dfrac{m_A v_A - m_A u_A}{t}}\\)........ [i] In the same way, Rate of change of momentum of B= \\({\dfrac{m_b v_B - m_B u_B}{t}}\\) Force exerted by A on B (\\(F_{ab}\\))= \\(F_{ab}= {\dfrac{m_B v_B - m_B u_B}{t}}\\).......... [ii] Newton's third law of motion states that every action has an equal and opposite reaction, then, \\(F_ab= -F_ba\\) [ ' -- ' sign is used to indicate that 1 object is moving in opposite direction after collision] Using [i] and [ii] , we have \\({\dfrac{m_B v_B - m_B u_B}{t}} = - {\dfrac{m_A v_A - m_A u_A}{t}}\\) \\(m_B v_B - m_B u_B= - m_A v_A + m_A u_A\\) Finally we get, \\(m_B v_B + m_A v_A = m_B u_B + m_A u_A\\) This is the derivation of conservation of linear momentum.

If a body of mass M collides against a wall with velocity V and rebounds with the same speed, then its change of momentum will be

Initial velocity of body before collision \\(u = -V\\)where minus sign indicates the body moves towards the wall.Initial momentum of the body \\(P_i = Mu = -MV\\)Final velocity of body after collision \\(v =V\\)where positive sign indicates the body moves away from the wall.Final momentum of the body \\(P_f = Mv = MV\\)Change in momentum \\(\Delta P = P_f - P_i = MV - (-MV) = 2MV\\)

Two masses of 1 gm and 4 gm are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is:

Kinetic energy \\(K = \dfrac{p^2}{2m}\\)Two masses are given as- \\(m_1 = 1 \ gm\\) and \\(m_2 = 4 \ gm\\)But \\(K_1 = K_2\\)\\(\therefore\\) \\( \dfrac{p^2_1}{2m_1} = \dfrac{P^2_2}{2m_2}\\)Or \\( \dfrac{P_1^2}{2(1)} = \dfrac{P^2_2}{2(4)}\\)\\(\implies\\) \\(P_1:P_2 = 1:2\\)

A bullet of mass \\(0.012\ kg\\) and horizontal speed \\(70\ ms^{-1}\\) strikes a block of wood of mass \\(0.4\ kg\\) and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Mass of the bullet, m = 0.012 kgInitial speed of the bullet, \\(u_b = 70 m/s\\)Mass of the wooden block, \\(M = 0.4 \ kg\\)Initial speed of the wooden block, \\(u_B\\) =0Final speed of the system of the bullet and the block = \\(v \ \ m/s\\)Applying the law of conservation of momentum:\\(m u_b + M u_B = (m + M) v \\)\\(0.012 \times 70 + 0.4 \times 0 = (0.012 + 0.4) v\\)\\(v = 0.84 / 0.412\\) \\( =2.04\\) m/sFor the system of the bullet and the wooden block:Mass of the system, \\(m' = 0.412 \ kg\\)Velocity of the system \\(=2.04 m/s\\)Height up to which the system rises = \\(h\\)Applying the law of conservation of energy to this system:Potential energy at the highest point \\(=\\) Kinetic energy at the lowest point\\(m'gh = (1/2)m'v^2\\)\\(h = \dfrac{v^2 }{2g}\\)\\(=\dfrac{ (2.04)^2}{2 \times 9.8} \\)\\(= 0.2123 m \\)The wooden block will rise to a height of \\(0.2123 m\\).The heat produced = Kinetic energy of the bullet - Kinetic energy of the system\\(= (1/2) mu^2 - (1/2) m'v^2\\) \\(=(1/2) \times 0.012 \times (70)^2 - (1/2) \times 0.412 \times (2.04)^2\\)\\(= 29.4 -0.857 = 28.54 J\\)

Linear Momentum of a System of Particles | Definition, Examples, Diagrams

definition

Linear Momentum

Linear momentum is a vector quantity defined as the product of an object's mass,

m

, and its velocity,

v

. Linear momentum is denoted by the letter

p

and is called momentum for short. Note that a body's momentum is always in the same direction as its velocity vector. The unit of momentum are kg. m/s.

p = m v

definition

Conservation of Linear Momentum

Conservation of linear momentum expresses the fact that a body or system of bodies in motion retains its total momentum, the product of mass and vector velocity, unless an external force is applied to it. In an isolated system (such as the universe), there are no external forces so momentum is always conserved.

F = \frac{d p}{d t}

Since

F = 0

Therefore

p =

constant

example

Explanation of Physical Phenomena using principle of conservation of Momentum

Example: Two balls of equal masses are thrown upwards along the same vertical line at an interval of

2

seconds with the same initial velocity of

39.2 m s^{- 1}

. Find the total time of flight of each ball, if they collide at a certain height, and the collision is perfectly inelastic.

Solution:
When the particles collide the height of both particles will be same,
For first particle time will be

t

and for second particle time will be

t - 2

h = u t - \frac{1}{2} g t^{2} = u (t - 2) - \frac{1}{2} g (t - 2)^{2}

u t - \frac{1}{2} g t^{2} = u t - 2 u - \frac{1}{2} g (t^{2} - 4 t + 4)

0 = - 2 u + 2 g t - 2 g

t = \frac{u}{g} + 1

t = 5 s

By solving the quadratic equation in

t

we get the time of first collision to be

5

h = 39.2 \times 5 - \frac{1}{2} g \times 5^{2} = 20 g - 12.5 g = 7.5 g m e t e r

Now by momentum conservation, velocity of particles after collision

m (u - g t) + m (u - g (t - 2)) = m v^{'}

m u - 5 m g + m u - 3 m g = m v^{'}

v^{'} = 2 u - 8 g = 0 m / s

a = - g

Now for collision with ground let time taken be

t_{2}

h = \frac{1}{2} g t_{2}^{2} \Rightarrow t = \frac{2 h}{g} = \frac{2 \times 7 . 5 g}{g} = 15

Thus, times of total time of flights will be

5 + 15

s and

3 + 15

law

Law of conservation of linear momentum

The law of conservation of linear momentum states that if no external forces act on a system, then the linear momentum of the system remains constant.

Linear Momentum of a System of Particles

definition

Linear Momentum

definition

Conservation of Linear Momentum

example

Explanation of Physical Phenomena using principle of conservation of Momentum

law

Law of conservation of linear momentum

A $5000 k g$ rocket is set for vertical firing. The exhaust speed is $800 m s^{- 1} .$ To give an upward acceleration of $20 m s^{- 2},$ the amount of gas ejected per second to supply the needed thrust is $(g = 10 m s^{- 2})$

A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 m/s. When the stone reaches the floor,the distance of the man above the floor will be

The force 'F' acting on a particle of mass 'm' is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to 8 s is

A block of metal weighing $2$ kg is resting on a frictionless plane it is struck by a jet releasing water at a rate of $1$ kg/s and at a speed

A particle moves in the x-y plane under the influence of a force such that its linear momentum is $p (t) = A (i cos (k t) - j sin (k t)]$ where A and k are constants. The angle between the force and
momentum is

State the law of conservation of momentum and derive related expression.

If a body of mass M collides against a wall with velocity V and rebounds with the same speed, then its change of momentum will be

Two masses of 1 gm and 4 gm are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is:

definition

Linear Momentum

definition

Conservation of Linear Momentum

example

Explanation of Physical Phenomena using principle of conservation of Momentum

law

Law of conservation of linear momentum

A 5000 kg rocket is set for vertical firing. The exhaust speed is 800 ms−1. To give an upward acceleration of 20 ms−2, the amount of gas ejected per second to supply the needed thrust is (g=10 ms−2)

A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 m/s. When the stone reaches the floor,the distance of the man above the floor will be

The force 'F' acting on a particle of mass 'm' is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to 8 s is

A block of metal weighing 2 kg is resting on a frictionless plane it is struck by a jet releasing water at a rate of 1 kg/s and at a speed

A particle moves in the x-y plane under the influence of a force such that its linear momentum is p​(t)=A(icos(kt)−jsin(kt)] where A and k are constants. The angle between the force and momentum is

Two sphere A and B of masses m1​ and m2​ respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity 2v​ in a direction perpendicular to the original direction. The mass A moves after collision in the direction.

State the law of conservation of momentum and derive related expression.

If a body of mass M collides against a wall with velocity V and rebounds with the same speed, then its change of momentum will be

Two masses of 1 gm and 4 gm are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is:

A $5000 k g$ rocket is set for vertical firing. The exhaust speed is $800 m s^{- 1} .$ To give an upward acceleration of $20 m s^{- 2},$ the amount of gas ejected per second to supply the needed thrust is $(g = 10 m s^{- 2})$

A block of metal weighing $2$ kg is resting on a frictionless plane it is struck by a jet releasing water at a rate of $1$ kg/s and at a speed

A particle moves in the x-y plane under the influence of a force such that its linear momentum is $p (t) = A (i cos (k t) - j sin (k t)]$ where A and k are constants. The angle between the force and
momentum is