JEE(Main), JEE(Advance) & CBSE Board

Center of mass, Momemtum, Collision

Introduction

Everybody in his/her childhood had played with “HIT-ME” at least once. Remember the toy, which no matter how much you punch it, never goes down! A truck however slow it may be, can cause casuality to any person to which it collides while a bicycle cannot cause much of a problem though it speed was quite high! Why does a rubber ball, a plastic ball, a super ball, all retract to different heights even though thrown from same height initially?

The Answer to all such queries lies in this chapter which opens new insight.

CENTER OF MASS:

It is the point in a system which behaves as though the entire mass of system is concentrated there and its motion is same if the resultant of all forces acting on the system were applied directly to it.

Mathematically:

Co-ordinates of Center of mass:

Where is total mass

Position Vector of Center of Mass:

Where are position vector of i^th particle.

Dumb Question:

1) Why does the Center of Mass divide internally the line joining two particles in inverse ratio of their masses?

Ans:

Consider two particles at O and P. Suppose Center of Mass is at C, Now fix origin at O. Then,

So,

This implies that the Center of Mass is nearer to the heavier particle (obviously) and so the distance is inversely related to mass!

For a continuous body, the formulae for Center of Mass becomes

A) Center of Mass of a uniform straight rod:

Derivation:

B) Center of Mass of semi circular wire:

Derivation:

Total length of semi circular wire =

Elemental length = Rdq

C) Center of Mass of a uniform semi circular plate:

Derivation:

Here the element chosen is a thin wire (semi circular) of radius r.

As derived earlier, the for this is at .

Dumb Question:

1) Why is here?

Ans: The mass dm is a semi circular thin wire whose position is variable (y is not unique), so we concentrate dm mass on the COM of this wire that is at .

D) Center of Mass of hemispherical shell:

Derivation:

Dumb Question:

1) Why is y = Rsinq here?

Ans: The Center of Mass of elemental ring is at its center (by symmetry) which is at a height of Rsinq from origin; hence y = Rsinq.

E) Center of Mass of a hemisphere:

Derivation:

Dumb Question:

1) Why is volume of elemental disc = Rdq (cosq) (pR²cos²q) and not

Rdq (pR²cos²q)?

Ans: This is a solid hemisphere. If we consider the curvature at the ends of the disc as negligible (as we do in a hollow shell as in the previous case) then the integration starts yielding wrong results. So as a tip we can take this that whenever the integration is being done over solid objects, the curvature effects cannot be neglected!

Motion of Center of Mass:

Differentiating with respect to time

The Center of mass of the system is not affected by the internal forces. If the external forces add up to zero, then center of mass has no acceleration.

Consider a system of n particles. Now the i^th particle will be acted by 2 forces- internal forces due to rest of the (n-1) particles and the external forces. So far the i^th particle,

= Force on i^th particle by j^th particle (j=i)

= Force on i^th particle by external forces.

Carrying summation over ‘i’ i.e. including all particles:

(Zero)

Dumb Question:

1) Why =0?

Ans: The expansion will look like this:

Now

Also

Hence

Hence if =0 then =0.

Illustration:

Find the acceleration of center of mass?

Assuming downwards as positive direction.

Illustration:

A projectile of mass 5m is projected from ground with velocity 50Ö2 at 45⁰ angle. At the highest point, it explodes into two pieces. One has a mass 4m while other has a mass m. Both pieces fly off in opposite horizontal direction. Mass 4m falls at a distance of 600m from point of projection. Find the distance of second mass from point of projection?

Solution:

Range of projectile otherwise;

So, if the distance of ‘m’ is x then,

Dumb Question:

1) Why Center of mass follow same path?

Ans: Now if there had been no explosion, the projectile would have landed at 500m from point of projection. Now even if the projectile has exploded, the center of mass must follow same path as no extra force is acting.

LINEAR MOMENTUM AND ITS CONSERVATION PRINCIPLE:

Linear momentum is defined as

For a system of n particles =

If m is constant,

Suppose =zero = = constant. This gives the principle of momentum conservation.

“The linear momentum of a system remains constant (both magnitude and direction) if external forces acting on the system adds up to zero”.

Illustration:

A man of mass 5kg climbs a rope of length 10m suspended below a balloon of mass 10kg. The balloon is stationary with respect to ground. If the man begins to climb up the rope at a speed 2m/s (relative to rope) in upward direction, what is the speed with which the balloon moves w.r.t ground?

Solution:

Given = 2 m/s (upward). Now take the balloon and man as one system.

If center of mass does not move =

(Downwards 2/3 m/s).

Dumb Question:

1) Why center of mass does not move here?

Ans: Since initially the balloon + Man system was stationary. So even if both move as no extra external force is acting (except gravity) over the center of mass. It continues to remain in its position. Hence center of mass does not move.

Illustration:

A gun of mass m₁ fires a bullet of mass hm₁ with speed V₀ relative to the barrel of gun. Which is inclined at an angle of a⁰ with horizontal. The gun is placed on smooth horizontal surface. Find racial speed of gun?

Solution:

So velocity components for bullet are:


V_rsinA

For bullet

(-V)

As no external force is acting over the system barrel + bullet. Hence we can apply principle of momentum conservation [in horizontal direction].

Dumb Question:

1) What is the speed of bullet w.r.t ground?

Ans:

Hence the speed as seen from ground will be addition of both speeds (barrel speed + its speed w.r.t barrel).

2) What happens to vertical direction momentum?

Ans: The vertical direction momentum = hmV_rsina is transferred to ground. The normal reaction N gives an impulse in normal direction to take care of this momentum.

V_rsin a

VARIABLE MASS:

These are the systems in which mass does not remain constant with respect to time.

Basic Equation:

Derivation:

The two states of the system are shown at instants t and t + Dt. Initially whole mass M moves with . After Dt time, DM moves with relative velocity while rest (M-DM) moves with . Note that absolute velocity of DM is .

On the system

Now,

Note: Here is negative, so thrust force becomes +ve.

Illustration: A rocket of initial mass m₀ (including shell and fuel) is fired vertically at time t = 0. The fuel is consumed at a rate of ‘a = At’ where A = constant, t = time. The gases are exhaust at a constant relative speed of ‘m’ m/s with respect to rocket. In addition to gravity, there is resistance from air which can be assumed to be a constant force of . Find the differential equation velocity of rocket at the time t?

Solution:

{This is the mass at time t}

Let the velocity of rocket at time ‘t’ = V.

This is the differential equation governing the velocity of rocket with respect to time.

Illustration:

The chain has mass per unit length = M kg/m. What should be the value of F so as to pull the chain with constant velocity v?

Solution:

Let at the time t

Dumb Question:

1) What is ?

Ans: If ‘dx’ length is also set into motion then mdx mass is set into motion.

2) What is the system here?

Ans: The system can be the chain that is being set into motion. Solving with time, the system is increasing in length and hence mass!

IMPULSUE:

Impulse is defined as the product of force and time. If force is constant then,

. Here time ‘t’ is the time for which force acts.

On the other hand if is not constant then where acts between time t₁ and t₂. This also represents the area under F-t graph with end points t₁ and t₂.

Dumb Question:

1) A man who falls from a height on a cement floor receives more injury than when he falls from same height on a heap of sand why?

Ans: Just before the collision the velocity of man is same in both cases. Now the momentum of man has to come to zero, so to change momentum either a large force can be applied in a short interval of time or a smaller force in a large interval of time. In case of cement does not deform. But sand deforms to same extent there by extending the time period. Hence force applied is lesser here.

COLLISION:

Collision can be defined as an event of short duration during which the colliding bodies exert large force on each other. During the collision the total momentum of system does not change. If no external forces are acting on the system. The impulsive forces acting during the collision are internal forces which have no effect on total momentum of system.

Collision – Elastic (Coefficient of restitution = 1) (No Loss of K.E)

- Inelastic (Coefficient of restitution <>

Collision - Head on (Directions of velocities are along the line of impact)

- Oblique (Otherwise)

Coefficient of restitution is defined as

=1 Perfectly Elastic Collision

=0 Perfectly Inelastic Collision

1) Head on Elastic Collision:

Derivation:

System = m₁+m₂; so momentum is conserved.

------- (1)

As elastic collision so (K.E)_initial= (K.E)_final

Note:

a) When i.e. velocities get exchanged.

b) If .

That is the smaller ball rebounds with same velocity while bigger one which was at rest remains at rest.

c) .

That is the bigger ball keeps on moving with same velocity while other moves with 2V₂.

2) Head on Inelastic Collision:

Derivation:

The two balls together form the system so linear momentum is still conserved.

------------ (1)

Now K.E doesn’t remain constant because there is loss of energy.

The second equation comes from

-------- (2)

Solving these two we get above results.

Illustration:

A ball of mass m₀ at a speed V₀ makes a head on collision with an identical ball at rest. KE of balls after the collision is of the original. Find the coefficient of restitution?

Solution:

OBLIQUE COLLISIONS:

When at least one body moves in a direction different from the line of impact, the collision is called oblique. In oblique collisions;

a) Impulses act along common normal and momentum is conserved in this direction.

b) No impulse act in tangential direction.

c) Definition of coefficient of restitution is applied only in normal direction.

d) Net momentum is conserved.

Illustration:

The ball of mass m₀ hits a floor with speed m at an angle q. The coefficient of restitution is . Find ‘f’ and velocity ‘v’?

Solution:

In n-n direction:

--------- (1)

In t-t direction:

mSinq = Vsinf -------- (2)

Solving (1) and (2) we get

Where .

Dumb Question:

1) Why the velocity component in t-t direction does does not change?

Ans: Because there is no impulse in t-t direction hence momentum remains

mSinq = Vsinf.

PROBLEMS (EASY TYPE)

1) A wooden plank of mass m₀ is resting on a smooth horizontal floor. A man of mass am₀ starts moving from one end of plank to other. The length of plank is l₀. Find the displacement of plank over the floor when man reaches other end of plank?

Solution:

The center of mass remains stationary as no external force. Hence,

2) A particle of mass 5kg is initially at rest. A force starts acting on it one direction whose magnitude changes with time. The force time graph is as shown in figure. Find the velocity of particle after 10 sec?

Solution:

Impulse = DMomentum

= Area under F-t graph.

3) Three identical balls, ball A, ball B, ball C are placed on a smooth floor on a straight line at the separation of 20m between balls as shown in figure. Initially balls are stationary. Ball A is given a velocity of 5m/s towards Ball B. Collision between A+B has coefficient of restitution as ½. But collision between B+C has coefficient of restitution as 1. What is the time interval between 2 consecutive collision between ball A and ball B?

Solution:

(1) Collision of A and B:

If velocities of ball A and ball B are V₁ and V₂ after collision then,

Then

And

(2) Collision between B and C:

Velocities get interchanged and so B is at rest after collision.

So time interval =

= 16 sec

4) A light spring of spring constant K₀ is kept compressed with compression =

between two masses of mass ‘m’ and ‘am’. When released the blocks acquire a velocity in opposite directions. The spring loses contact with both blocks when it comes in natural length! Find the final velocities of the two blocks?

Solution:

Since no external force is acting, hence

But initial momentum = 0. Hence if final velocities of m and am are V₁ and V₂ then,

mV₁ - amV₂ = 0

V₁ = aV₂ ------ (1)

Now energy remains conserved so,

E_i = E_f.

5) A bullet of mass m₀ strikes a block of mass ‘hm₀’ with a speed of V₀ and gets embedded into the mass ‘hm₀’. The block is attached to a spring of stiffness ‘K’. Find the loss of K.E. of system after impact?

Solution:

As the spring force is a non impulsive force, so we can conserve momentum as:

6) A stationary body explodes into four identical fragments such that three of them fly off mutually perpendicular to each other with same K.E =E₀ Find energy of explosion?

Solution:

Let the three fragments move along x, y, z direction with velocities .

Now

7) A system of two blocks A and B are connected by an inextensible mass less string as shown. Pulley is mass less and frictionless. A bullet of mass m₀ with velocity V₀ and gets embedded. Find the impulse imparted by tension force?

Solution:

Let the Impulse =

Then for Bullet + B Þ

8) After perfectly inelastic collision between 2 identical particles moving with same speed in different direction, the speed of particles between ¾ th of initial speed. Find the angle between the two before collision.