Centre of Mass, Momentum and Collision Concepts Page - 4

Definition

Thrust Force

Example: Sand drop from a stationary hopper at a rate dmdt${\displaystyle \frac{dm}{dt}}$ on to a conveyor belt moving with a velocity v$v$ in the reference frame of the laboratory. Since dvdt=0${\displaystyle \frac{dv}{dt}}=0$, Therefore, it is an example of variable mass. To an observer at rest on the belt the falling sand would appear to have a horizontal motion with speed v$v$ in a direction opposite to that shown for the belt in the laboratory. Therefore vrel=−v.$v_{rel}=-v.$ In this example dmdt${\displaystyle \frac{dm}{dt}}$ is positive as the system is gaining mass. Note that in the absence of friction mass of the belt does not enter the problem. power supplied by the external force is P=F¯¯¯¯⋅v¯¯¯=v¯¯¯⋅(v¯¯¯dmdt)=v2dmdt$P=\overline{F}\cdot \overline{v}=\overline{v}\cdot (\overline{v}{\displaystyle \frac{dm}{dt}})=v^2{\displaystyle \frac{dm}{dt}}$. 

Solution:
Since KE=12mv2$KE={\displaystyle \frac{1}{2}}m{v}^2$ and d(KE)dt=v22dmdt${\displaystyle \frac{d(KE)}{dt}}={{\displaystyle \frac{v}{2}}}^2{\displaystyle \frac{dm}{dt}}$.
Also, P=v2dmdt=2d(KE)dt$P=v^2{\displaystyle \frac{dm}{dt}}=2{\displaystyle \frac{d(KE)}{dt}}$

Definition

Elastic Collision

Definition: An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms.

Example:
Collision between two elastic marbles.

Example

Inelastic Collision

An inelastic collision, in contrast to an elastic collision, is a collision in which kinetic energy is not conserved due to the action of internal friction. In collisions of macroscopic bodies, all kinetic energy is turned into vibrational energy of the atoms, causing a heating effect, and the bodies are deformed.

Example: In a bouncing ball each impact of the ball is inelastic, meaning that energy dissipates at each bounce.

Example

Velocity of approach and velocity of separation for particles colliding for inelastic collision

When the two bodies stick together after collision, then it is perfectly inelastic collision and in this case, the coefficient of restitution e$e$ is equal to zero.

 In perfectly inelastic collision, the two colliding bodies gets stuck together and move with a common velocity.For perfectly inelastic collision, e=0$e=0$.
In perfectly inelastic collision, total momentum of the system remains conserved just before and just after the collision but some kinetic energy is lost during the collision.

Definition

Force during collision

In a car accident, the metal body of the car gets dented. The force acting on the car during the accident changes the shape of the car.

Example

Problem based on conservation of momentum in elastic collision

Example: A ball of mass M$M$ moving with a velocity V$V$ collides head on elastically with another of same mass but moving with a velocity v$v$ in the opposite direction. What will be the scenario after collision?

Solution:
Using conservation of momentum we have
M1u1−M2u2=−M1v1+M2v2$M_1{u}_1-M_2{u}_2=-M_1{v}_1+M_2{v}_2$or
MV−Mu2=−Mv+Mv2$MV-M{u}_2=-Mv+M{v}_2$
or
V−u2=−v+v2$V-u_2=-v+v_2$.............(i)
and as for a perfect elastic collision we have e=1, thus we get
v2+v=u2+V$v_2+v=u_2+V$.................(ii)
Using these two equations we get 
u2=v$u_2=v$ and v2=V$v_2=V$
Thus the velocities are exchanged between the two balls.

Example

Perfectly inelastic Collisions

A perfectly inelastic collision occurs when the maximum amount of kinetic energy of a system is lost. In a perfectly inelastic collision, i.e., a zero coefficient of restitution, the colliding particles stick together. In such a collision, kinetic energy is lost by bonding the two bodies together. This bonding energy usually results in a maximum kinetic energy loss of the system.

Physical Example: Sticking of mud to the wall.

Definition

Conservation of Momentum

Momentum of a system is constant, if there are no external forces acting on the system. For a collision occurring between object 1 and object 2 in an Isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

Example

Conservation of kinetic energy in elastic collision

Example: A plank of mass M$M$ is placed on a smooth horizontal surface. Two light identical springs, each of stiffness K$K$, are rigidly connected to struts at the end of the plank as shown in the figure. When the spring is in their unextended position, the distance between their free ends is 3$3$l$l$. A block of mass m$m$ is placed on the plank and pressed against one of the springs so that it is compressed to l$l$. To keep the block at rest it is connected to the strut by means of a light string. Initially, the system is at rest. Now the string is burnt. Find the maximum kinetic energy.
Solution:
Using Conservation of energy and conservation of momentum, the equations we get are:
Kl22=12mv2+12MV2${\displaystyle \frac{K{l}^2}{2}}={\displaystyle \frac{1}{2}}m{v}^2+{\displaystyle \frac{1}{2}}M{V}^2$andmv+MV=0$mv+MV=0$
where v=$v=$ max velocity of m$m$
and V=$V=$ max velocity of M$M$
Solving these two equations:
we get:
v=KMm(M+m)−−−−−−−−−−√l$v=\sqrt{{\displaystyle \frac{KM}{m(M+m)}}}l$
KEmax=mv22=KMl22(M+m)${KE}_{max}={\displaystyle \frac{m{v}^2}{2}}={\displaystyle \frac{KM{l}^2}{2(M+m)}}$

Example

Conservation of Energy during collision

Example: A ball of mass m$m$ moving with speed v$v$ undergoes a head-on elastic collision with a ball of mass nm$nm$ initially at rest. What is the fraction of the incident energy transferred to the second ball?

Solution:Given :         
m1=m$m_1=m$                
m2=nm$m_2=nm$Let the velocities of  A$A$ and B$B$ after the collision be  v1$v_1$ and  v2$v_2$  respectively.initial velocity of A before collision is  v$v$ and that of B is zero.Initial kinetic energy of the system         E=12mv2$E={\displaystyle \frac{1}{2}}m{v}^2$
Using         v2=(m2−em1)u2+(1+e)m1u1m1+m2$v_2={\displaystyle \frac{(m_2-e{m}_1)u_2+(1+e)m_1{u}_1}{m_1+m_2}}$  v2=(nm−1×m)(0)+(1+1)m(v)m+nm$v_2={\displaystyle \frac{(nm-1\times m)(0)+(1+1)m(v)}{m+nm}}$       ⟹v2=2v(1+n)$⟹v_2={\displaystyle \frac{2v}{(1+n)}}$Thus Kinetic energy of B after the collision         E′B=12(nm)v22$E_B^{\prime }={\displaystyle \frac{1}{2}}(nm)v_2^2$
E′B=12(nm)4v2(1+n)2$E_B^{\prime }={\displaystyle \frac{1}{2}}(nm){\displaystyle \frac{4v^2}{(1+n{)}^2}}$      
Fraction of total kinetic energy retained by B              E′BE=4n(n+1)2${\displaystyle \frac{E_B^{\prime }}{E}}={\displaystyle \frac{4n}{(n+1{)}^2}}$   

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