Mod 8

Module 8: Elastic and inelastic Collision Momentum and Impulse: Let us consider a particle of mass m. We can write Newton’s second law of motion for this particle as

  d  dp ∑ F = dt [mv ] = dt , ……………. [1] The net force [vector sum of all forces] acting on a particle equals the time rate of change of momentum of the particle. This, not

 p

  F ∑ = ma , is the form in which Newton originally stated his second law of motion.

It is valid only in inertial frames of reference. To see the physical difference between momentum and kinetic energy, we first define a quantity closely related to momentum called impulse. Let us first consider a particle acted on by a constant net force



 F ∑ during a time interval ∆t from t1 to t2.

The impulse of the net force, denoted by I , is defined to be the product of the net force and the time interval:

   I = ∑ F[ t 2 − t 1 ] = ∑ F∆t , ……….. [2] Impulse is a vector quantity; its direction is the same as the net force

 F ∑ .

   dp dp If the net force ∑ F is constant, then is also constant. In that case, is equal to the total change in dt dt   momentum p 2 − p 1 during the time interval t2-t1, divided by the interval:  p 2 − p1 ∑F = t − t 2 1 Multiplying this equation by [t2 – t1], we have

   F ∑ [t 2 − t 1 ] = p 2 − p1 Comparing to Eq. [2], we end up with a result called the impulse-momentum theorem:

   I = p 2 − p 1 , ………….. [3] The change in momentum of a body during a time interval equals the impulse of the net force that acts on the body during that interval.

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The impulse-momentum theorem also holds when forces are not constant. To see this, we integrate both

 dp sides of Newton’s second law ∑ F = over time between the limits t1 and t2: dt t2

 F ∫ ∑ dt = t1

  p2    dp ∫t dt dt = p∫ dp = p 2 − p 1 1 1

t2

The integral on the left is defined to be the impulse

 I=

t2

 I of the net force



∑ F during this interval:



∫ ∑ Fdt , …………… [4]

t1

Equation [4] is the general definition of impulse.

We can define an average net force

 Fav such that even when

  F ∑ is not constant, the impulse I is given

by

  I = Fav [ t 2 − t 1 ] , ……………. [5] When

   F F is constant, ∑ ∑ = Fav and Eq. [5] reduces to Eq. [2].

Linear Momentum: Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion. The amount of momentum which an object has is dependent upon two variables: how much stuff is moving and how fast the stuff is moving. Momentum depends upon the variables mass and velocity. Momentum = mass * velocity p=m*v where m = mass and v=velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity. The units for momentum would be mass units times velocity units. The standard metric unit of momentum is the kg*m/s.

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Momentum is a vector quantity. The direction of the momentum vector is the same as the direction of the velocity of the ball. In a previous unit, it was said that the direction of the velocity vector is the same as the direction which an object is moving.

The Law of Linear Momentum Conservation: One of the most powerful laws in physics is the law of momentum conservation. The law of momentum conservation can be stated as follows. 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. That is, the momentum lost by object 1 is equal to the momentum gained by object 2. The above statement tells us that the total momentum of a collection of objects (a system) is "conserved" that is the total amount of momentum is a constant or unchanging value. Consider a collision between two objects - object 1 and object 2. For such a collision, the forces acting between the two objects are equal in magnitude and opposite in direction (Newton's third law). This statement can be expressed in equation form as follows.

The forces act between the two objects for a given amount of time. Regardless of how long the time is, it can be said that the time that the force acts upon object 1 is equal to the time that the force acts upon object 2. As an equation, this can be stated as

Since the forces between the two objects are equal in magnitude and opposite in direction, and since the times for which these forces act are equal in magnitude, it follows that the impulses experienced by the two objects are also equal in magnitude and opposite in direction. As an equation, this can be stated as

But the impulse experienced by an object is equal to the change in momentum of that object (the impulsemomentum change theorem). Thus, since each object experiences equal and opposite impulses, it follows logically that they must also experience equal and opposite momentum changes. As an equation, this can be stated as

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The above equation is one statement of the law of momentum conservation. In a collision, the momentum change of object 1 is equal and opposite to the momentum change of object 2. That is, the momentum lost by object 1 is equal to the momentum gained by object 2. In a collision between two objects, one object slows down and loses momentum while the other object speeds up and gains momentum. If object 1 loses 75 units of momentum, then object 2 gains 75 units of momentum. Yet, the total momentum of the two objects (object 1 plus object 2) is the same before the collision as it is after the collision; the total momentum of the system (the collection of two objects) is conserved.

Collision: A collision is an isolated event in which two or more bodies (the colliding bodies) exert relatively strong forces on each other for a relatively short time.

Before

During

After

Figure 1: A flowchart showing the system in which a collision occurs.

Collisions can be

(i) (ii)

elastic, or totally inelastic.

(i) Elastic Collision: In an elastic collision energy and momentum are conserved. An elastic collision is a collision in which the total kinetic energy of the colliding bodies after collision is equal to their total kinetic energy before collision. Elastic collisions occur only if there is no conversion of kinetic energy into other forms, as in the collision of atoms (Rutherford backscattering is one example.) In the case of macroscopic bodies this will not be the case as some of the energy will become heat. In a collision between polyatomic molecules, some kinetic energy may be converted into vibrational and rotational energy of the molecules, but otherwise molecular collisions appear to be elastic. Example: In billiards, collisions play an important role. Because the collisions between billiard balls are almost perfectly elastic, and the balls roll on a low-friction surface, their predictable behaviour is often used to illustrate Newton's laws of motion.

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(ii) Inelastic Collision: Inelastic collision is a collision in which some of the kinetic energy of the colliding bodies is converted into internal energy in one body so that kinetic energy is not conserved. In collisions of macroscopic bodies some kinetic energy is turned into vibrational energy of the atoms, causing a heating effect. Collisions between molecules of a gas or liquid may also be inelastic as they cause changes in vibrational and rotational energy levels. An extreme case is the perfectly inelastic collision, in which all of the kinetic energy relative to the centre of mass is converted to thermal or internal energy of the system, and the two objects stick together after the collision and move as one body. After a perfectly inelastic collision, however, both bodies have the same velocity; the sum of their kinetic energies is reduced, compared with the initial value, because a part of it has changed into internal energy (warming up). In nuclear physics, an inelastic collision is one in which the incoming particle causes the nucleus it strikes to become excited or to break up.

Elastic Collisions in one dimension: (1) Stationary Target: Consider a simple head-on collision of two bodies of masses m1 and m2 as shown in Fig. 2. For convenience, we take one of the bodies to be stationary, with velocity v 2i = 0 before the collision. That body will be the target, and the other body will be the projectile, with velocity v 1i before the collision. We assume that this two-body system is closed (no mass enters or leaves it) and isolated (no net external force acts on it). Let us also make another special another assumption: the kinetic energy of the system is the same before and after the collision. The collision is then said to be of a special type called elastic collisions.

v1i

m1

V2i= 0

m2

(a) Before

VCM

m1 m2

v1f

m1

(b) During

V2f

m2

(c) After

Figure 2: Two bodies undergo an elastic collision. One of them (the target body with mass m2) is initially at rest before the collision. The velocities are shown (a) before, (b) during and (c) after the collision. The conservations of linear momentum and of kinetic energy for the collision of Fig.2. gives us

m1v 1i = m1v 1f + m 2v 2f and 1 1 1 m1v 12i = m 1v 12f + m 2v 22f 2 2 2

(linear momentum) - - - - - - - - - - - - - - (1)

(kinetic energy)

5

--------------

(2)

In each of these equations, the subscript i identifies the initial velocities and the subscript f the final velocities of the body. If we know the masses of the bodies and if we also know v 1i, the initial velocity of body 1, the only unknown quantities are v1f and v2f, the final velocities of the two bodies. Equation (1) can be rewrite

m1 (v 1i − v 1f ) = m 2v 2f

- - - - - - - - - - - (3)

And Eq. (2) can be rewrite

m1 (v 12i − v 12f ) = m 2v 22f m1 (v 1i − v 1f )(v 1i + v 1f ) = m 2 v 22f After dividing Eq.(4) by Eq. (3)

m1 (v 1i − v 1f )(v 1i + v 1f ) m1 (v 1i − v 1f )

=

m 2 v 22f m 2 v 2f

(v 1i + v 1f ) = v 2f − − − − − − − − − −(5) Putting the value of v2f in Eq. (3) we get

m1 (v 1i − v 1f ) = m 2 (v 1i + v 1f ) v 1i (m1 − m 2 ) = v 1f (m1 + m 2 ) v 1f =

(m 1 − m 2 ) v 1i − − − − − − − − − − (6) (m 1 + m 2 )

And putting the value of v1f in Eq. (5) we get

v 1i +

(m 1 − m 2 ) v 1i = v 2f (m 1 + m 2 )

(m 1 + m 2 + m 1 − m 2 ) v 1i = v 2f (m 1 + m 2 ) v 2f =

2m 1 v 1i − − − − − − − − − − − (7) (m 1 + m 2 )

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- - - - - - - - - - - (4)

That is v2f is always positive, that is the target body with mass m 2 always moves forward. From Eq. (6) we see that v1f may be of either sign that is the projectile body with mass m1 moves forward if m 1 > m2 but rebounds if m1 < m2.

Some special situation: (a) Equal masses: If m1 = m2, then Eq. (6) and (7) becomes, v1f = 0

and

v2f = v1i

That is after a head-on collision of bodies with equal masses, body 1 (projectile) stops and body 2 (target) takes off with the initial speed of body 1. In head-on collisions, bodies of equal masses simply exchange velocities. This is true even if the target particle is not initially at rest.

(b) A massive target: A massive target means, m2 >> m1, Then Eq. (6) and (7) becomes v1f = -v1i (neglecting m1)

 2m 

1 v 1i − − − − − − − − − − − − − (8) and v 2f ≈  m  2 

That is body 1 simply bounces back in the same direction from which it came, its speed essentially unchanged. Body 2 moves forward with a low speed, because the quantity in parentheses in Eq. (8) is much less than unity.

(c) A massive projectile: That is m1 >> m2. The n the Eq. (6) and (7) becomes v1f ≈ v1i and

v2f ≈ 2v1i (neglecting m2) ----------------------------------------- (9)

Equation (9) tells us that body 1 simply keeps on going, scarcely slowed by the collision. Body 2 charges ahead at twice the speed of the body 1.

(2) Moving Target: Let us consider the elastic collision in which both bodies are moving before they undergo an elastic collision.

v1i

v2i x

m1

7

m2

Fig. 3: Two bodies headed for an elastic collision.

For the situation of Fig. (3), the conservation of linear momentum is written as

m1v 1i + m 2v 2i = m1v 1f + m 2v 2f − − − − − − − − − − − − − − − (10) And the conservation of kinetic energy is written as

1 1 1 1 m1v 12i + m 2v 22i = m1v 12f + m 2 v 22f − − − − − − − − − − − −(11) 2 2 2 2 Equation (10) and (11) can be written as

m1 (v 1i − v 1f ) = − m 2 (v 2i − v 2f ) − − − − − − − − − − − − − − − − − (12) and m1 (v 1i − v 1f )(v 1i + v 1f ) = − m 2 (v 2 i − v 2f )(v 2i + v 2f ) − − − − − − − (13) After dividing Eq. (!3) by Eq.(12) we get,

m1 (v 1i − v 1f )(v 1i + v 1f ) − m 2 (v 2i − v 2f )(v 2 i + v 2f ) = m1 (v 1i − v 1f ) − m 2 (v 2i − v 2f ) (v 1i + v 1f ) = (v 2 i + v 2f ) v 2f = v 1i + v 1f − v 2 i − − − − − − − − − − − − − − − − − − − − − − − − (14) Putting the value of v2f in Eq. (12) we get

m1v 1i − m 1v 1f = − m 2v 2i + m 2 v 1i + m 2v 1f − m 2v 2 i

v 1i ( m1 − m 2 ) = v 1f ( m1 + m 2 ) − 2m 2 v 2 i then, v 1f =

( m1 − m 2 ) 2m 2 v 1i + v ( m1 + m 2 ) ( m1 + m2 ) 2i

− − − − − − − − − − − − − − − − (15 )

Putting the value of v1f in Eq. (14) we get

( m1 − m2 ) 2m 2 v 1i + v −v ( m1 + m2 ) ( m1 + m 2 ) 2i 2 i ( m 1 + m 2 + m1 − m 2 ) 2m 2 − m 1 − m 2 = v 1i + v 2i ( m1 + m 2 ) ( m1 + m2 )

v 2f = v 1i +

∴ v 2f =

2m 1 m 2 − m1 v 1i + v − − − − − − − − − − − − − − − (16) ( m1 + m 2 ) ( m1 + m 2 ) 2i

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Inelastic Collisions in one dimension: Figure 4 shows a one dimensional inelastic collision in which one body is initially stationary. The law of conservation of linear momentum holds, so

m1v = (m1 + m 2 )V − − − − − − − − − − − − (17) or V =

m1 v − − − − − − − − − − − −(18) (m 1 + m 2 ) v

m1

V

m2 at rest

m1 + m2

Before

After

Fig. 4: A completely inelastic collision between two bodies.

Where V represents the final velocity of the stuck-together bodies. Equation (18) tells us that the final speed is always less than that of the incoming body. If both bodies (projectile and target) are moving prior to a completely inelastic collision, we replace Eq. (17) with

m1v 1 + m 2 v 2 = (m 1 + m 2 )V − − − − − − − − − − − − − − − −(19) or V =

m1 m2 v1 + v 2 − − − − − − − − − − − − (20) (m 1 + m 2 ) (m 1 + m 2 )

In Eq. (19) m1v1 is the initial linear momentum of one body, and m2v2 is that of the other body.

COMPLETELY INELASTIC COLLISIONS: Let us look at what happens to momentum and kinetic energy in a completely inelastic collision of two bodies [A and B]. Because the two bodies stick together after the collision, their final velocities must be equal:

   v A 2 = v B2 = v 2 , …………[1] Conservation of momentum gives the relation

   m A v A1 + mB v B1 = [m A + mB ]v 2 , ………[2]

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Suppose, for example, that a body with mass inelastically with a body of mass component of velocity

m A and initial x-component of velocity v 1 collides

mB that is initially at rest [ v B1 = 0 ]. From Eq. [2] the common x-

v 2 of both bodies after the collision is v2 =

mA v 1 , ………………[3] m A + mB

The kinetic energies K1 and K2 before and after the collision, respectively, are

K1 =

mA 1 1 1 2 ]2 v 12 . m A v 12 and K 2 = [m A + mB ] v 2 = [m A + mB ][ 2 2 m + m 2 A B

Therefore, the ratio of final to initial kinetic energy is

K2 mA = , …………[4] K 1 m A + mB The right hand side is always less than unity and, therefore, the kinetic energy after a completely inelastic collision is always less than before.

Elastic collisions in two dimensions: We consider a glancing collision (it is not head-on) between a projectile body and a target body at rest. Figure 5 shows a typical situation. After the collision, the two bodies fly off at angles θ1 and θ2, as the figure shows. y

V2f

θ1

m2 V1i

x θ2

m1

V1f

Fig. 5: An elastic collision between two bodies in which the collision is not head-on. The body with mass m2 (the target) is initially at rest.

From the conservation of linear momentum (a vector relation), we can write two scalar equations:

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m1v 1i = m 1v 1f cos θ 1 + m 2 v 2f cos θ 2

(x - component) - - - - - - - - - - - - - (21)

and 0 = - m1v 1f sin θ 1 + m 2 v 2f sin θ 2

(y - component) - - - - - - - - - - - - - -(22)

If the collision is elastic, kinetic energy is also conserved:

1 1 1 m1v 12i = m 2v 12f + m 2v 22f 2 2 2

(kinetic energy) - - - - - - - - - - - - - - - (23)

These three equations contain seven variables: two masses, m1 and m2; three speeds, v1i, v1f, and v2f and two angles, θ1 and θ2. If we know any four of these quantities, we can solve the three equations for the remaining three quantities.

Solved Examples: Example 1: Suppose you through a ball with a mass of 0.40 kg against a brick wall. It hits the wall moving horizontally to the left at 30 m/s and rebounds horizontally to the right at 20 m/s. [a] Find the impulse of the net force on the ball during its collision with the wall. [b] If the ball is in contact with the wall for 0.010 s, find the average horizontal force that the wall exerts on the ball during the impact. Solution: [a] The initial x-component of momentum of the ball is p1 = mv1= [0.40 kg][-30 m/s] = - 12 kg. m/s The final x-component of momentum is p2 = mv2 = [0.40 kg][20 m/s] = 8.0 kg. m/s The change in the x-component of momentum is p2 – p1 = mv2 – mv1 = 8-[-12] = 20 kg. m/s. Therefore, The x-component of impulse of the net force on the ball = Ix =mv2x – mv1x = 20 kg. m/s = 20 N.s. [b] The average horizontal force that the wall exerts on the ball during the impact = [ Fav ]x = Ix / ∆t = 20 N.s/0.010s = 2000 N. Example 2: A soccer ball has a mass of 0.40 kg. Initially, it is moving to the left at 20 m/s, but then it is kicked and given a velocity at 450 upward and to the right, with a velocity of magnitude of 30 m/s. Find the impulse of the net force and the average net force, assuming a collision time of 0.010 s. Solution: v1x = - 20 m/s, v1y = 0 v2x = v2y = [30 m/s] [0.707] = 21.2 m/s, since sin 450 = cos 450 = 0.707 Ix = p2x – p1x = m[v2x – v1x] = [0.40 kg][21.2 – (-20)] = 16.5 kg.m/s Iy = m[v2y – v1y] = [0.40 kg][21.2 – 0] = 8.5 kg.m/s The components of the average net force on the ball are [Fav]x = Ix/∆t = 1650 N, [Fav]y = Iy /∆t =850 N The magnitude and direction of the average force are

(1650) 2 + (850) 2 = 1.9x10 3 N − 1 850 ] = 27 0 where Θ is measured from the +x-axis. And Θ = tan [ 1650 Fav =

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Example 3: A marksman holds a rifle of mass mR = 3.00 kg loosely in his hands, so as to let it recoil freely when fired. He fires a bullet of mass mB = 5.00 g horizontally with a velocity relative to the ground of vB = 300 m/s. What is the recoil velocity vR of the rifle? What are the final momentum and kinetic energy of the bullet? Of the rifle? Solution: Conservation of the x-component of total momentum gives

Px = 0 = mB v B + mR v R ,

vR = −

mB 0.005 v B = −[ ][300 m / s ] = −0.50 m / s mR 3.00

The momentum and kinetic energy of the bullet are

PB = mB v B = 0.005 x300 = 1.5 kg.m / s , K B =

1 1 mB v B2 = 0.005 x[300 ]2 = 225 J 2 2

The momentum and kinetic energy of the rifle are

PR = mR v R = [ 3.00 ][ −0.50 ] = −1.5 kg.m / s , K R =

1 1 mR v R2 = [ 3.00 ][ −0.50 ]2 = 0.375 J 2 2

Example 4: A thorium-227 nucleus at rest decays into a radium-223 nucleus [mass 223 u] by emitting an alpha particle [mass 4 u]. The kinetic energy of the alpha particle is found to be 6.00 MeV. What is the kinetic energy of the recoiling radium nucleus? Solution:

1 2 mRa v Ra . 2 From conservation of momentum, we get m α v α = mRa v Ra . The kinetic energy of the radium nucleus =

K Ra =

Therefore, we can write

K Ra =

m 1 1 4 mRa [m α v α mRa ]2 = α [ m α v 2α ] = [6.00 MeV ] = 0.107 MeV 2 mRa 2 223

Example 5: A small car of mass 1.2x103 kg traveling east at 60 km/h collides at an intersection with a truck of mass 3.0x103 kg traveling north at 40 km/h. The car and truck stick together after the collision. Find the velocity of the wreckage just after the collision. Solution:

 P are = 1.2x10 3 kgx16.667m / s + 3.0x10 3 x0 = 20.00004 x10 3 kg.m / s

The components of the total momentum

Px = Pcx + Ptx = m c v cx + m t v tx

Py = Pcy + Pty = 1.2x10 3 kgx0 + 3.0x10 3 x11.111 = 33.333 x10 3 kg.m / s P = [20.00004 x10 3 ]2 + [33.333 x10 3 ]2 = 38.87 x10 3 kg.m / s tan Θ =

Py Px

= 1.6666,

Θ = 59.035 0 P v = = 38.87 x10 3 / 4.2x10 3 = 9.25 m / s M

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Problem Sheet - Collision 1.

A 2000 kg car moving 25 m/s runs head-on into a 1500 kg car initially at rest. If the collision is perfectly inelastic, find (a) each car’s speed after collision, and (b) the ratio of the system’s final kinetic energy to its initial kinetic energy. (Ans: 14.3 m/s, (b) 0.57)

2.

In a feat of public marksmanship, you fire a bullet into a hanging target. The target, with bullet embedded, swing upward. Noting the height reached at the top of the swing, you immediately inform the crowd of the bullet’s speed. For arbitrary masses m 1 and m2, and height h, how would you calculate the speed? (a) If the mass of the bullet is 12 g, the mass of the block on the ballistic pendulum (hanging target) is 2kg, and the final height is 10.4 cm, what speed did you announce to the crowd? (Ans: 240m/s)

3.

A 4-kg block moving right at 6m/s collides elastically with a 2-kg block moving right at 3 m/s. Find their final velocities. (Ans: 4 m/s and 7 m/s)

4.

A neutron of mass m1 and speed v1i collides elastically with a carbon nucleus of mass m2 at rest. (a) What are the final velocities of both particles? (b) What fraction of its initial energy does the neutron lose? Ans: (a) -0.845 v1i , 0.154 v1i (b)]

5.

A 2-kg box moving at 3 m/s makes an elastic collision with a stationary 4-kg box. (a) What is the original mechanical energy? (b) How much energy is transferred to the 4-kg box? [Ans: (a) 9J (b) 8J]

6.

Two metal spheres, suspended by vertical cords, initially just touch. Sphere 1, with mass m1= 30g, is pulled to the left to height h1 = 8.0 cm, and then released. After swing down, it undergoes an elastic collision with sphere 2, whose mass m2 = 75 g. (a) What is the velocity v1f of sphere 1 just after the collision? (b) To what height h1’ does sphere 1 swing to the left after the collision? (c) What is the velocity v2f of sphere 2 just after the collision? (d) To what height h2 does sphere 2 swing after the collision? (Ans: (a) -0.54 m/s (b) 1.5 cm, (c) 0.72 m/s (d) 2.6 cm)

7.

A Karate expert strikes downward with his fist (of mass m1 = 0.70 kg), breaking a 0.14 kg board. He then does the same to a 3.2 kg concrete block. The spring constant k for bending are 4.1x10 4 N/m for the board and 2.6x106 N/m for the block. Breaking occurs at a deflection d of 16 mm for the board and 1.1 mm for the block. (a) Just before the board and block break, what is the energy stored in each? (b) What fist speed v is required to break the board and the block? Assume that the mechanical energy is converted during the bending, that the fist-object collision at the onset of bending is completely inelastic. (Ans: for board 5.2J and block 1.6J (b) 4.2 m/s and 5.0 m/s).

8.

Two skaters collide and embrace, in a completely inelastic collision. That is, they stick together after impact suggested in Fig, where the origin is placed at the point of collision. Alfred, whose mass mA is 83 kg, is originally moving east with speed vA = 6.2 km/h. Barbara, whose mass mB is 55 kg, is originally moving north with speed vB = 7.8 km/h. (a) What is the V of the couple after impact? (b) What is the fractional change in the kinetic energy of the skaters because of the collision? (Ans: (a) 40o and 4.9 km/h, (b) -0.5).

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Y (north)

v

vA

θ

M = mA+mB x (east)

mA CM

VB