Centre of Mass, Momentum and Collision Concepts Page - 3

Definition

Centre of Gravity of irregular lamina

Take a lamina with three holes near the periphery of the lamina, now  suspend the lamina through them, one by one. Draw a line of equilibrium for each suspension point. The point of intersection of these three lines would be the centre of gravity.

Definition

Identification of centre of gravity

A solid body can be balanced by supporting it at its centre of gravity. For example, a uniform metre rule has its centre of gravity at the mark 50cm. It is possible because the algebraic sum of moments of the weights of all particles of rule about any point is zero. This point where the rule is supported is its centre of gravity.

Definition

Linear Momentum

Linear momentum is a vector quantity defined as the product of an object's mass, m$m$, and its velocity, v$v$. Linear momentum is denoted by the letter p$p$ 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.
p⃗ =mv⃗ $\overrightarrow{p}=m\overrightarrow{v}$

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.

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=dp⃗ dt$F={\displaystyle \frac{d\overrightarrow{p}}{dt}}$
Since F=0$F=0$
Therefore p=$p=$ 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$2$ seconds with the same initial velocity of 39.2ms−1$39.2m{s}^{-1}$. 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$t$ and for second particle time will be t−2$t-2$
h=ut−12gt2=u(t−2)−12g(t−2)2$h=ut-{\displaystyle \frac{1}{2}}g{t}^2=u(t-2)-{\displaystyle \frac{1}{2}}g{(t-2)}^2$
ut−12gt2=ut−2u−12g(t2−4t+4)$ut-{\displaystyle \frac{1}{2}}g{t}^2=ut-2u-{\displaystyle \frac{1}{2}}g(t^2-4t+4)$
0=−2u+2gt−2g$0=-2u+2gt-2g$
t=ug+1$t={\displaystyle \frac{u}{g}}+1$
t=5s$t=5s$
By solving the quadratic equation in t$t$ we get the time of first collision to be 5$5$s.
h=39.2×5−12g×52=20g−12.5g=7.5g meter$h=39.2\times 5-{\displaystyle \frac{1}{2}}g\times 5^2=20g-12.5g=7.5gmeter$
Now by momentum conservation, velocity of particles after collision
m(u−gt)+m(u−g(t−2))=mv′$m(u-gt)+m(u-g(t-2))=m{v}^{\prime }$
mu−5mg+mu−3mg=mv′$mu-5mg+mu-3mg=m{v}^{\prime }$
v′=2u−8g=0m/s$v^{\prime }=2u-8g=0m/s$
a=−g$a=-g$
Now for collision with ground let time taken be t2$t_2$
h=12gt22⇒t=2hg−−−√=2×7.5gg−−−−−−−√=15−−√$h={\displaystyle \frac{1}{2}}g{t}_2^2\Rightarrow t=\sqrt{{\displaystyle \frac{2h}{g}}}=\sqrt{{\displaystyle \frac{2\times 7.5g}{g}}}=\sqrt{15}$
Thus, times of total time of flights will be 5+15−−√$5+\sqrt{15}$s and  3+15−−√$3+\sqrt{15}$s

Definition

Reduced Mass and Problem involving reduced mass

Definition: When two bodies in relative motion are acted upon by a central force involving Newton's law then the system can be replaced by a single mass called the reduced mass.

Example: A light spring of force constant K$K$ is held between two blocks of masses m$m$ and 2m$2m$. The two blocks and the spring system rests on a smooth horizontal floor. Now the blocks are moved towards each other compressing the spring by x$x$ and then they are suddenly released. Then find the relative velocity between the blocks when the spring attains its natural length.

Solution:Potential energy of the spring is being converted into kinetic energy of system.
Potential energy of the spring before released =12Kx2$={\displaystyle \frac{1}{2}}K{x}^2$
K.E of the system in Center of mass frame =12μv2relative$={\displaystyle \frac{1}{2}}\mu v_{relative}^2$ , where μ=m×2mm+2m=2m3$\mu ={\displaystyle \frac{m\times 2m}{m+2m}}={\displaystyle \frac{2m}{3}}$
is reduced mass of the system.

Now, 12μv2relative=12Kx2${\displaystyle \frac{1}{2}}\mu v_{relative}^2=\frac{1}{2}K{x}^2$
v2relative=32mKx2$v_{relative}^2={\displaystyle \frac{3}{2m}}K{x}^2$
vrelative=(3K2m−−−−√)x$v_{relative}=(\sqrt{{\displaystyle \frac{3K}{2m}}})x$

Definition

Define reduced mass and use in solving problems in two dimensions

In a collision with a coefficient of restitution e$e$, the change in kinetic energy can be written as:
ΔK=12μv2rel(e2−1)$\mathrm{\Delta }K={\displaystyle \frac{1}{2}}\mu v_{rel}^2(e^2-1)$
where vrel$v_{rel}$ is the relative velocity of the bodies before collision and μ$\mu$ is the reduced mass of the system.

Example

Example on momentum conservation in two blocks attached by a spring

Two blocks of masses m$m$ and 3m$3m$ are placed on a friction less, horizontal surface. A light spring is attached to the more massive block, and the
blocks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after which the block of mass 3m$3m$ moves to the right with a speed of V⃗ 3m=2.60i⃗  m/s${\overrightarrow{V}}_{3m}=2.60\overrightarrow{i}m/s$.)What is the velocity of the block of mass m$m$? (Assume right is positive and left is negative.)
Momentum is conserved and we have :
mv⃗ m+3mv⃗ 3m=0⃗ $m{\overrightarrow{v}}_m+3m{\overrightarrow{v}}_{3m}=\overrightarrow{0}$
v⃗ m=−3mmv⃗ 3m${\overrightarrow{v}}_m={\displaystyle \frac{-3m}{m}}{\overrightarrow{v}}_{3m}$
=−3v⃗ 3m$=-3{\overrightarrow{v}}_{3m}$
=−3×2.60i⃗ $=-3\times 2.60\overrightarrow{i}$
=−7.80i⃗ $=-7.80\overrightarrow{i}$
|v⃗ m|=7.80m/s$|{\overrightarrow{v}}_m|=7.80m/s$ 

Example

Problems on phenomenon related to Rocket Propulsion

Example: Like rocket propulsion problem of variable mass can think of rain drops falling through a cloud of small water droplets. Some of these small droplets adhere to the rain drop. Thereby increasing its mass as it falls. The force on the rain drops is 
Fext=dpdt=mdvdt+vdmdt$F_{ext}={\displaystyle \frac{dp}{dt}}={\displaystyle \frac{mdv}{dt}}+{\displaystyle \frac{vdm}{dt}}$
Assume the mass of the raindrop depends on the distance x$x$ that it has fallen. Then m=kx$m=kx$, where k$k$ is a constant and 
dmdt=kv${\displaystyle \frac{dm}{dt}}=kv$, This gives since Fext=mg$F_{ext}=mg$
mg=mdvdt+v(kv)$mg={\displaystyle \frac{mdv}{dt}}+v(kv)$
dividing by k$k$, we get xg=xdvdt+v2$xg={\displaystyle \frac{xdv}{dt}}+v^2$ where x=mk$x={\displaystyle \frac{m}{k}}$
This is a differential equation that has a solution of the form v=at$v=at$, where a$a$ is acceleration and is constant. Take initial velocity of the raindrops to zero. Then find the value of a$a$.
Solution:
xdvdt+v2=gx${\displaystyle \frac{xdv}{dt}}+v^2=gx$ may be written as dvdt+v2x=g${\displaystyle \frac{dv}{dt}}+{\displaystyle \frac{v^2}{x}}=g$ 
Its solution is ve∫dtt=∫ge∫dttdt$v{e}^{\int {\displaystyle \frac{dt}{t}}}=\int g{e}^{\int {\displaystyle \frac{dt}{t}}}dt$

Solution to this differential equation is:
vt=gt22$vt={\displaystyle \frac{g{t}^2}{2}}$
v=at$v=at$ we get a=g2

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