One Dimensional Elastic Collision

Collision Momentum

One Dimensional Elastic Collision

Now, we are going to discuss the application of the law of momentum conservation on the collision of two bodies. Collision process may happen very shortly, for example on the collision between two billiard balls, or it may happen in a long time, for example on the collision among stars in heaven. On all collision processes, the colliding bodies will interact with each other strongly only during the collision. If there are external forces, these forces are much smaller compared to the interacting forces and therefore are neglected.

If the total kinetic energy of the colliding bodies after the collision is equal to the total kinetic energy before the collision, the collision process is called completely elastic collision. However, if the total kinetic energy of the two bodies before the collision is not the same with the total kinetic energy after the collision, then the collision process is called inelastic collision. In an inelastic collision, part the of the kinetic energy is changed into another kind of energy, for example, thermal energy. Thus, the total kinetic energy after the collision the two bodies are joined together, the collision process is called completely inelastic collision, as studied in Example 4.2. It should be underlined here that in every collision process where the influence of external forces is neglected, the law of momentum conservation always holds.

1. One-Dimensional Elastic Collision

I have been mentioned that if in a collision there is no lost of kinetic energy, then a collision is elastic. We will discuss a one-dimensional collision, in which the velocities of colliding bodies is on the same straight line, for example along the x-axis.

Figure 4.6 (a) shows two bodies of masses m1 and m2, which are moving with velocities of v1 and v2 along the same straight line. Having collided elastically, their velocities become v1' and v2' (Figure 4.6 (b)). Based on the law of momentum conservation, it is obtained:

2. Two-Dimensional Elastic Collision

If often happens that the motion of colliding bodies is not in a straight line but in the two-dimensional plane. One example of such collisions in the collision between two balls in a billiards game.

Figure 4.7 shows a ball of mass m1 moving along the x-axis is colliding with another ball of mass m2, which in initially at rest. Having colliding, the two ball move into directions that make the angles of Ө₁ and Ө₂ respectively toward the x-axis.

By applying the law of momentum conservation in the x-axis direction, it is obtained

m₁v₁ = m₁v₁’ cos Ө₁ + m₂v₂’ cos Ө₂

 Since initially both bodies are not moving in the direction of the y-axis then the component of momentum in the y-axis directions is zero.

m₁v₁’ sin Ө₁ + m₂v₂’ sin Ө₂ = 0

3. Completely Inelastic Collision

In a completely inelastic collision, after the collision, the two bodies will move together so that it holds v₁’ = v₂’= v'. Thus, the law of momentum conservation becomes:

m₁v₁ + m₂v_{2 = (} m_{1 +} m₂) v’

Therefore, the velocities of the two bodies after the collision can be calculated with the formula. Thus, by measuring the masses and velocities of the bodies before the collision, the velocity after collision could be predicted. In a complete collision, kinetic energy after collision is always smaller than kinetic energy before collision. If kinetic energy before the collision is E_k = ½ m₁V₁'². The kinetic energy after collision is

E_k' = ½ (m₁ + m₂) v’²

4. Inelastic collision

Most collisions between two bodies found in nature are inelastic collisions, for example, a tennis ball colliding with racket or the hit ball in a baseball game. The analysis of inelastic collision involves a quantity called restitution coefficient (e).

Restitution coefficient is defined as the negative value ofthe ratio of relative velocities before and after collision. 

For inelastic collision, the value of restitution coefficient is between zero and one, 0 < e < 1.

Exercise 4.3 

1. A body with the mass of 1 kg is moving to the direction of positive x-axis with a velocity of 2 m/s. The other body which mass is 3 kg, is moving in the opposite direction with a velocity of 2 m/s. After collided, both bodies are moving together as one. Calculate the velocity of both bodies after the collision.
2. Two balls with masses of 3 kg and 6 kg are moving with speeds of 4 m/s and 1 m/s respectively. The two balls are approaching each other along a straight line connecting the center of masses of the balls. After the collision, the ball with a mass of 3 kg stops moving. Determine the value of the restitution coefficient of the two balls.
3. Two bodies A and B have the same masses. Initially, A is moving to the right with an initial velocity of 5 m/s, and after 2 seconds A covers a distance of 14 m. At that time, A and B undergo a completely inelastic collision. If B is initially moving to the left with a velocity of 15 m/s, determine the velocity of both after the collision.

Multiple Choice

1) Andre Agassi is playing tennis using a ball with the mass of 100 g. The ball is approaching him with a velocity of 100 m/s. Andre hits back the ball with a force of 120 N. If the racket touches the ball for 0.02 s, the velocity of the ball is .....
a. 46 kg.m/s
b. 56 kg.m/s
c. 66 kg.m/s
d. 76 kg.m/s
e 86 kg.m/s

2) Among the following moving bodies, the one that experiences the greatest force when the body hits the wall and stops in the same interval of time is ...........
a. 40 kg with a speed of 25 m/s
b. 50 kg with a speed of 15 m/s
c. 100 kg with a speed of 10 m/s
d. 150 kg with a speed of 7 m/s
e. 200 kg with a speed of 5 m/s

3) Someone with the mass of 40 kg is standing on a skateboard, which mass is 2 kg and is moving at a speed of 10 m/s. If he is suddenly jumping forward with the velocity of 4 m/s while the skate board keeps running, the magnitude of the skate board's velocity is ............
a. 120 kg.m/s
b. 130 kg.m/s
c. 135 kg.m/s
d. 140 kg.m/s
e 150 kg.m/s

4) Ball A is moving with momentum mv and colliding with ball B which is moving on the same straight line. If after the collision the momentum of ball A become -3 mv, the increase of the momentum of ball B is ..............
a.  2 mv
b.  -2 mv
c.  3 mv
d.  -4 mv
e.  4 mv

5) A bullet with the mass of 10 g is fired and hits a block which masses 1.49 kg. The block hanging freely on a piece of rope which length is 0.2 m. If the acceleration due to the earth's gravity is 10 m/s² and the rope is displaced by 60 cm measured from its initial position/vertical position, the velocity of the bullet when fired is ........... m/s
a.  300 √3
b.  300 √2
c.  150 √3
d.  50 √2
e.  150

6) A grenade that is initially at rest is suddenly exploding and breaking up into two pieces, which are moving in the opposite directions. The ratio of masses of the two pieces m₁ : m₂ = 1 : 2. If the released energy is 3 x 10⁵ joules, the ratio of the kinetic energy of the first piece and the second one is ..........
a. 1 : 1
b. 2 : 1
c. 1 : 3
d. 5 : 1
e. 7 : 5

7)  A ping pong ball is falling freely from the height of h₁ and reflected back with the lower height of h₂. The restitution coefficient is ..........
a.   e = ( h₂/ h₁ )^1/2
b.   e = ( h₁/ h₂ )^1/2
c.   e = ( h₁/ h₂ )
d.   e = ( h₂/ h₁ )
e.   e = h₁ h₂

Conservation of Momentum

Momentum

Conservation of Momentum

Conservation of Momentum

Show two balls with masses of m₁ and m₂ are moving in opposite directions in a straight line with masses of m1 and m2 are moving in opposite directions in straight line with velocities of v₁ and v₂, respectively. Having collided, their velocities become v₁' a v₂'.

The sum of momentum of both balls before collision is

p = p₁ + p₂  = m₁v₁ + m₂v₂ 

The sum of momentum of both balls after collision is

p' = p₁' + p₂'  = m₁v₁' + m₂v₂' 

Based on the law of momentum conservation p=p'. thus

m₁v₁ + m₂v₂  = m₁v₁' + m₂v₂' 

The law of momentum conservation holds only if the sum of external forces working on the colliding bodies is equal to zero.

Example 4.2

A trained coach with a mass of 10.000 kg is running with the speed of 24 m/s to the right. The train coach collides with a similar coach which is initially at rest. If the two coaches are then connected as the result of the collision, what is their velocity after the collision?

Answer :
The sum of momentum before collision is
 p₂  = m₁v₁ + m₂v₂ = (10.000 kg) (24 m/s) + (10.000 kg) ( 0 m/s)

       = 2,4 x 10⁵ kg.m/s   (the direction of the initial total momentum is to the right)

The law of momentum conservation

Having collided, the two coaches merge and move together with velocity v'. the sum of momentum before the collision is equal to the sum of momentum after the collision and the momentum becomes:

 p₂  = (m₁ + m₂)v'

Based on the law of momentum conservation, the sum of momentum before collision is equal to the sum of momentum after collision;  p₂  = p₁    
(m₁ + m₂)v' = p₁  ⇒   v'  =  p₁ /(m₁ + m₂)
                     = 2,4 x 10⁵ kg.m/s (10.000 kg + 10.000 kg) = 12m/s.
After a collision, both coaches are moving together to the right with a velocity of 12 m/s.

Exercise 4.2

1) The velocity of a bullet when it is fired is 200 m/s. If the masses of the bullet and the rifle are 10 g and 5 kg respectively, then what is the recoil velocity of the rifle toward the shoulder of the man when the bullet is fired?2. Bodies A and B have masses of 3 kg and 2) kg respectively. Body A is moving to the right with velocity of 5 m/s and B is moving to the left with velocity of 10 m/s in such a way that they are colliding with each other. If after the collision the two bodies are joined together then what is the velocity of the two bodies after collision.

Working Principles of a Rocket

There is wrong assumption that the forward movement of a rocket is due to the reaction of the thrust face done by the rocket to the air. However, this concept does not satisfy when the rocket is in outer space since there is no air to push the rocket.

The working principle of a rocket is similar to the working principle of a rifle firing a bullet. Consider that the rifle in a position to move freely without friction as illustrated in Figure 4.8. When the bullet is fired, the rifle will move backward with a certain velocity. In this situation, the law of momentum conservation holds.

The acceleration experienced by the rifle is similar to the acceleration acquired by the rocket. The acceleration of the is obtained from the thrust done by the rocket. Every molecule of gas can be assumed as on small bullet fired by the rocket. In that system, the total momentum of the rocket is always equal to the total momentum of the gas; as long as there is no external force that influences the rocket (the external forces that influence the rocket are neglected).

The force of gravity, which is acting as the external force, will decrease the momentum of the rocket (remember, a force can change the momentum of a body). Consider that the initial velocity of the rocket is v and its mass is m. The rocket ejects gas with the mass of Δm and velocity v_g so that the velocity of the rocket eject increases to v + Δv. The velocities of rocket and gas are measured relatively toward the same frame of reference, for example; toward the earth. Based on the values of velocities, the initial and final momentum of the rocket could be calculated as follows ( p₁ and p₂, respectively);

P₁ = mvP₂ = (m – Δm) (v + Δv)

The final momentum of the gas is p_g = - Δmv_g, where the negative sign shows that the gas velocity is opposite the rocket velocity.
Base on the law of momentum conservation, it is obtained

 mv = (m – Δm) (v + Δv) – Δmv_g

mv = mv – Δmv­ + m Δv – ΔmΔv - Δmv_g  

0   = – Δmv­ + m Δv - Δmv_g
The value Δm.Δv is neglected since it is much smaller compared to the other terms, thus we obtain the change of the rocket's velocity as follows

Exercise 4.4

1. Determine the escape velocity needed by a rocket to leave the earth's surface. (g = 9.8 m/s², earth's radius is 6,400 km).
2. A rocket is flying with the rate of gas ejection of 300 kg/s. If the mass of the rocket is 450 kg and its relative velocity toward the gas is 300 km/hours, determine the average acceleration of the rocket.
3. A block with a mass of 1.5 kg is placed on a horizontal plane. The friction coefficient of the block with the horizontal plane is 0.2 A bullet with a mass of 10 g is fired horizontally, hits the block and buried in it. The block is displaced by 1 m. If g = 10 m/s², then determine the speed of the bullet when it collides with the block.
4. Block A which mass is 3 kg, is at rest on a floor. The kinetic friction coefficient of the floor is 0.2. Block A is collided by block B, which mass is 2 kg and moving at the speed of 20 m/s. After the collision, block A and B are moving together. Determine the distance covered before both of them stop.
5. A ball with the mass of 60 g is colliding with the wall perpendicularly. The ball is then turning back with the speed of 40 m/s, which is the same speed as the ball's speed before it collided with the wall. What is the momentum change experienced by the ball during the collision?
6. A bullet with 10 gram of mass hits the block with 1.5 kg of mass on a surface of a horizontal plane, with the friction coefficient of 0.2. Due to the shot, the block is displaced by 1-meter. Determine the speed of the bullet just before it hits the block. 

Multiple Choice

1)  A bullet with 10-gram mass is fired at the velocity of 300 m/s to 1.49 kg mass of block which is freely hung on a 1.6-meter long rope. If the earth's gravitational acceleration is 10 m/s², the rope will deflect from the vertical position with an angle of...

a.  30^o
b.  37^o
c.  45^o
d.  60^o
e.  67^o

2)  A 200 gram mass of marble is rolling on a frictionless floor so that its momentum is 4 kg.m/s. The marble's speed is....

a.  5 m/s

b.  10 m/s

c.  15 m/s
d.  20 m/s
e.  25 m/s

3)  A 100 gram mass of the ball, which slides at speed of  100 m/s is struck back so that its speed becomes 76 m/s. If the duration of the contact between the ball and the stick is 0.02 second, the force done by the stick is ...........
a.  100 N
b.  120 N
c.  176 N
d.  24 N
e.  2 N

4)  A car with 800 kg mass, which moves at speed 2 m/s, is braked by friction force of 32 N. The distance covered by the braking car, before it finally stops, is....
a.  50 m
b.  25 m
c.  15 m
d. 5 m
e. 10 m

5)  A 2 kg particle moves in a straight path at positive x-axis direction with the velocity of 3 m/s. If a 6 N force has been working on the particle for 3 s, after 3 s ...........
a.  its speed will be 3 m/s
b.  its velocity direction is toward the negative x-axis
c.  the particle never stops
d.  its velocity keeps on increasing
e.  the direction of the particles is unchanged

You Need to Know

The Function of Helmet

One of the equipment of a motor rider is the helmet. The inner layer of a helmet is made of a soft material such as foam rubber sponge. This layer has a function to lengthen the time of impact to the head so that it reduces the contact force to the head during the shock. What happens if the helmet does not have soft layer? When shock happens, the interaction force with the head will take place shortly and it can cause a headache.

Impulse and Momentum Theorem

Momentum momentum and impluse physics

Impulse and Momentum Theorem

The Meaning of linear momentum and Impulse. Rocket launching is a spectacular example of the application of the law of momentum conservation. The rocket gains propulsion by burning and ejecting its fuel's mass. At a certain hight, the rocket also gains velocity increase by releasing its fuel tanks. What is momentum? what is its relationship with mass and velocity

When a billiards game is started, the white ball is shot hard so that is moves and collides with the other set of balls. The other balls the move to several directions with different velocities. Some are moving to the right, left, slanting to the right, to the left, etc. some are moving fast, some are moving slowly, some even stay at rest. Why?

What are the factors made the balls to move in different directions and at different speeds? To answer such question, we should learn about linear momentum, Impulse, and the law of momentum conservation.

Impulse and Momentum Theorem

Linear momentum (often simply called momentum) of a body is defined as the product of the body's mass and its velocity. momentum, symbolized by p, is a vector quantity. Mathematically, it is written

P = mv,

where m is the mass of the body and v is the velocity.

The SI unit of momentum is kg.m/s. the direction of momentum is the same as the direction of its velocity.

In daily lives, according to equation (4.1), a body with a certain mass, which is moving at high velocity, will have greater momentum than another body with the same mass but is moving at a lower speed. On the other hand, two bodies that are moving with the same velocity but having different masses will also have different momentums. For example, a container truck will have a greater momentum than a sedan, even though they both are moving with the same velocity.

If force F is applied to a body with mass m, then Newton's second law holds F = ma. However,
a = Δv/Δt  so that we get;
Where Δp represents a change in momentum.
Equation (4.2) is another statement of the Newton's second law are have discussed before. It states that.

The force working on a body equal to the change of momentum per unit interval of time.

Equation [4.2] can be applied in the case where the mass of the body is changing toward the time, for example; in a motion of a rocket, which loses it mass when the fuel is burning out.
We could apply the concept of momentum change to a body where the applied force is not constant. For example is momentum in a ball, which is rolling at a certain velocity, and then kicked off in such a way that its velocity changing from to. In this situation, Equation (4.2) could be used to determine the average force working on the ball, according to the formula:

The equation above can be expressed as

P Δt = ΔP = P_t - P₀ = mv_t - mv

The product between F  and Δt is called impulse, denoted by I. Thus,

I = ΔP = P₁ - P₀ 

From its definition, the impulse has a unit of N.s, however, based on Equation (4.4), the unit of impulse is the same with the unit of momentum, since impulse is a momentum change. Thus, N.s = kg.m/s

Example 4.1

1. A wagon with 15 kg of mass moves at velocity of 0,25 m/s toward a concrete wall. At the same time, a bullet with 10 g mass is fired at velocity 1.000 m/s toward the wall. Calculate : (a)  the momentum of the wagon, and (b)  the momentum of bullet.

Answer :
m_wagon = 15 kg;  v_wagon = 0,25 m/s
m_bullet = 10 g = 10 x 10^-3 kg;   v_bullet = 1.000 m/s

(a) the momentum of the wagon

P_wagon = m_wagon  v_wagon
           = (15 kg) (0,25 m/s)
           = 3,75 kg.m/s

(b) the momentum of the bullet

P_wagon = m_wagon  v_wagon
           = (10^-2 kg) (1,000 m/s)
           = 10 kg.m/s


2.  A circus player who has 70 kg of mass is falling freely from the height of 3 m. (a) Calculate the impulse experienced by the circus player. (b) Determine the average force given by his leg when he is landing with the bending knee so that distance covered, which is measured from the first time he is touching the ground until completely stop, is 50 cm.

Exercise 4.1

1. A ball is initially at rest. The ball is then hit by a stick so that it is moving with the velocity of 10 m/s. If the mass of the ball is 250 g; calculate the impulse applied to the ball
2. A force of 70 N is applied to a body for 0.8 seconds. Determine the impulse experienced by the body.
3. A horizontal force of 200 N is working on a body, which mass is 2 kg. If the body is initially at rest and the force is working for 0.4 s, what is the final velocity of the body?
4. A stone with a mass of 500 g is thrown up vertically upward (g = 10 m/s² ). At the highest point, the stone is breaking up into two pieces with the mass ratio 2:1. If the two pieces are moving n opposite directions, determine the ratio of their kinetic energies.

Multiple Choice

1)  Impulse is classified into.......... quantity

a.  vector

b.  scalar

c.  derived and scalar

d.  fundamental

e.  vector and derived

2) The dimension of impulse is ......

a.   MLT

b.   MLT^-1

c.   ML²T

d.   MLT^-2

e.   ML²T^-2

3) A body that has a mass of 20 kg is hit by a force of 60 N. If the force is working for 0.2 s, the impulse working on the body is .............
a. 9 Ns
b. 10 Ns
c. 11 Ns
d. 12 Ns
e. 13 NS

4) A marble with mass of 200 g is sliding on a slipery floor with velocity of 20 m/s. The magnitude of momentum of the marble is ............
a. 2 kg.m/s
b. 3 kg.m/s
c. 4 kg.m/s
d. 5 kg.m/s
e 6 kg.m/s

5) An apple with mass of 200 g is falling down from a 5 m high tree, (g = 10 m/s² ). The momentum of the apple when it hits the ground is ............
a. 12 kg.m/s
b. 10 kg.m/s
c. 9 kg.m/s
d. 8 kg.m/s
e 7 kg.m/s

6)   A bullet with a mass of 8 gram is fired to a block with a mass of 250 gram, which is initially at rest on the edge of a table with the height of 1 meter. The bullet is buried into the block and after the collision, the block landed on the floor at a distance of 10-meter measure form the legs of the table. The initial velocity of the bullet is......
A)  224 m/s
B)  184 m/s
C)  145 m/s
D)  42 m/s
E)  86 m/s

7)  In a collision process, the following statement that always holds is ...........
A)  the law of momentum conservation
B)  The law of kinetic energy conservation
C)  the law of mechanical energy conservation 

D)  the law of momentum conservation and the law kinetic energy conservationE)  the law of momentum conservation and the low of mechanical energy conservation.

8)  A body with the mass of 2.5 kg is falling freely from the height of 3 meters above the floor (g = 10 m/s² ). If the body undergoes a completely inelastic collision with the floor, then the heat produced is ...
A)  7.5 cal
B)  18 cal
C)  30 kcal
D)  75 cal
E)  300 cal

9)  A body with mass of 0.1 kg is moving to the west with velocity of 0.2 m/s, and having a completely elastic collision with another body with mass of 0.15 kg, which is initially at rest. The velocity of the 0.10 kg body after collision is ......
A)  0.16 m/s to the east
B)  0.16 m/s to the west
C) 0.08 m/s to the west
B) 0.04 m/s to the east
E) 0.04 m/s to the west

10)  Two balls with masses of m1 = 20 kg and m2 = 25 kg respectively are moving into the same direction on a straight line. Ball 1 is moving with velocity of 8 m/s situated behind ball 2, which is moving with velocity of 4 m/s. If both balls are colliding inelastically with e = 0.3, the magnitude of velocities of the two balls respectively are....
a.  6.31 m/s and 5.11 m/s
b.  6.41 m/s and 5.11 m/s
c.  6.31 m/s and 5.31 m/s
d.  6.51 m/s and 5.11 m/s
e.  6.31 m/s and 5.41 m/s