Chapter 7

Impulse and Linear Momentum

In outer space where there is no friction to support motion, if the cord that attaches a repairman to his space station is detached, the space man cannot walk back to the station unless he throws the wrench he has, for example, in opposite direction.  If the purpose is to move himself to the left, he has to throw his wrench to the right.  If his mass (including space suit and other attachments) is 100kg and the wrench mass is 1.00 kg, and he is able to throw the wrench at 25.0m/s to the left, then the recoil of his action makes him move to the right at 0.250m/s.  This means that the product Mv to the right equals the negative of product Mv to the left.  This is the way our Universe functions.  The conclusion is that the product Mv (a vector quantity) is an important quantity in physics that allows us analyze a number of physical phenomena.

Linear Momentum is defined as the "product of mass and linear velocity."  Pushing an object to the right,  results in a recoil (reaction) to the left.  A rifle (attached to a cart) fired to the left makes the rifle move to the right as shown:

It is easy to verify this concept both mathematically and experimentally.

Mb vb - Mrvr.   or   (Momentum to the left  = - Momentum to the right).

Momentum is a Vector Quantity:

Mv is a vector, because velocity is a vector.  This requires momentum to have direction.


Example 1:  A solid ball of mass M1 = 0.15kg is rolling to the right at v1 = 4.0m/s and another ball of mass M2= 0.35kg is rolling to the left at v2 = 6.0m/s.  Find (a) the momentum of each ball and (b) the net momentum.


(a) M1v1= 0.15kg(4.0m/s) = 0.60 kg m/s  ;  M2v2= 0.35kg(-6.0m/s) = -2.1 kg m/s.

(b)  ΣM= M1v1 + M2v2 = -1.5 kg m/s    (The net momentum is to the left).

Example 2:  A baseball of mass 0.120kg served by a pitcher horizontally to the left at 17m/s returns to the right at 63m/s after getting struck by a bat.  Calculate the change in its momentum.

Solution: Since Δ denotes a change and means a final value minus its corresponding initial value, we need to calculate Δ(Mv) = Mvf - Mvi .

Δ(Mv) = M(vf - vi = 0.120kg[ (+ 63m/s) - (-17m/s) ] = + 9.6 kg m/s.



The impulse (I) is defined as the "product of a force and a time interval." Mathematically, impulse is shown as FΔt.   For example, if a grocery cart is pushed with a constant force of 44N to the left for 25 seconds, the impulse of the pusher on the cart is

I = FΔt  ;  I = (-44N)(25s) = -1100 Ns.


Equivalence of Impulse and Linear Momentum:

1) Unit-wise:  It is easy to show that Ns is equal to kg m/s.

To show that Ns = kg m/s,   replace N by its equivalent: kgm/s2, to get:  

Ns = (kg m/s2)(s)  =  kg m/s.  Write with horizontal fraction bars to see it better.


2) Quantity-wise: 

It is also easy to verify that  FΔt  and  Δ(Mv)  are equivalent.  This means that "impulse" and "momentum" are equivalent.  In fact, we can show that:

Impulse = Change in Momentum.

Proof:  Make sure to write all steps down using horizontal fraction bars.

Use F = Ma;  replace a by Δv/Δt;  multiply through by Δt;  replace Δv by vf - vi to get:

        F = MΔv/Δt   ;   FΔt = MΔv   ;    FΔt = M(vf - vi)   or,    

FΔt  = Mvf - M vi .

Interpretation: This equation states that "The impulse of force F on mass M during Δt is equal to the change in the momentum of mass M."

Example 3:  A stationary train car of mass 12,000kg gets hit by another car moving to the right and is pushed with an average force of 4500N for a period of 4.2s.  Find the final velocity of the stationary car.

Solution:  Using the equivalence of impulse and linear momentum, results in:

FΔt = M (vf - vi ;  +4500N(4.2s) = 12,000kg(vf - 0 ;  vf  = +1.6 m/s  (to the right).

Example 4:  A 0.150-kg base ball is thrown horizontally to the left by a pitcher.  Its velocity just before getting hit by the bat is 15m/s to the left and after the strike becomes 45m/s to the right.  Find (a) the change in velocity Δv,  (b) the change in momentum MΔv, (c) the impulse of the bat on the ball, and (d) the average force of the bat on the ball if the contact time is 0.020s.

Solution: (a) Δv = vf - vi  = +45m/-(-15m/s) = + 60 m/s. 

 (b) MΔv = 0.150 kg(+ 60 m/s) = +9.0 kg m/s.

 (c) Based on the equivalence of impulse and momentum,  FΔt = 9.0 kg m/s as well.

                (d) FΔt = +9.0 kg m/s   ;   F(0.020s) = +9.0 kg m/s   ;   F = +450N.


Conservation of Linear Momentum:

When a system of particles go through collisions with each other, the total or net momentum remains constant.  A good example is the gas molecules in a closed solid cylinder that is thermally isolated from the ambient.  If no heat energy is given to it or taken from it, since its volume, pressure, and temperature remain constant, we may think that the net or total momentum of its many many molecules must also remain constant.  But we know that those trillion trillions of gas molecules in the cylinder keep colliding with each other in a random manner and in every collision there is momentum transfer.  The transfers are in between those isolated molecules and the inner walls of the cylinder.

The goal of this topic is to prove that the net or total momentum transfer in between the particles in an isolated system is constant.  We will prove this for the simplest possible case: the head-on collision of just two particles. 

Head-on Collision:  When the centers of two uniform spherical particles are moving exactly along the same line, their collision is called the "head-on" collision.   Let us consider the head-on collision of only two billiard balls that move toward each other along the same straight line.  Also suppose that both balls have uniform densities throughout and have exactly the same size (not necessarily the same mass).  That way their centers can move exactly along the same line as they roll and the collision will be a head-on and not oblique. The following figure shows solid spheres A and B with masses M1 and M2 that move at velocities v1 and v2 toward each other along the same line.  There are 3 stages: before collision, during collision, and after collision, as shown:

Let the velocities before collision be V1 and V2, and after collision be U1 and U2.

According to Newton's 3rd law, during collision, the impulse of M2 on M1 must be equal to the negative of the impulse of M1 on M2.

Total momentum before collision is:   M1V1 + M2V2 

Total momentum after collision is  M1U1 + M2U2   

During collision, each ball acts as a wall for the other.  In fact, each ball acts as a baseball bat for the other and imparts an impulse on the other ball.

According to Newton's 3rd law, the impulse of ball A on ball B must be equal to the  impulse of ball B on ball A, but in opposite direction.   One impulse is FABΔt, and the other -FBAΔt.   Forces are equal, and the contact times Δt is the same for both. 

We can write:    FABΔt = M2U2 - M2V2    and     FBAΔt = M1U1 - M1V1.

Since  FABΔt = - FBAΔt ; therefore,     M2U2 - M2V2 =  - (M1U1 - M1V1).

Rearranging yields:    

M1U1 +  M2U2  =  M1V1  +  M2V2.

This simply shows that

Total momentum after collision  =  Total momentum before collision.

 in other words, linear momentum is conserved.


Example 5:  A 1.00-kg toy car moving to the right at 1.40 m/s is hit from behind with a 0.500-kg piece of dough thrown horizontally also to the right at 3.60 m/s that causes the car and dough combination move faster.  Calculate the speed of the car-dough combo, knowing that the dough sticks to car.

Solution:  Let vcd be the velocity of the car-dough combo.  Total momentum after collision must be equal to the total momentum before collision.  This results in:

Mcvc + Mdvd  =  (Mc + Md) vcd    ;   

(1.00kg)(1.40m/s) + (0.500kg)(3.6m/s) = (1.00 + 0.500)kg(Vcd)   ;   vcd = 2.13 m/s.


Example 6: A 4.50-kg rifle is fixed on a 1.50-kg cart so that its barrel points horizontally to the right.  The cart can roll with negligible friction and is initially at rest.  The rifle is fired with a remote control device and shoots a 45.0 gram bullet to the right  As a result the rifle-cart set moves initially to the left at 2.50 m/s.  Calculate the bullet exit speed.

Solution:  Total momentum after collision must be equal to the total momentum before collision.  Since before firing (or collision), both the bullet and rifle are at rest, total momentum before firing is zero.  According to the law of conservation of  linear momentum, the total momentum after firing must also be equal to zero. This means that

      Mbvb    Mrvr      =     Mbvb      +    Mrvr     ;      (Mr includes rifle and cart).

0.045kg(0) + 6.00kg(0) = 0.045kg(vb)  + 6.00kg(-2.50m/s)   or,

        0       +     0         = 0.045kg(vb)   -   15.0 kg m/s          or,

                15.0 kg m/  = 0.045kg vb           or,     vb  = +333 m/s.   


Elastic and Inelastic Collisions:

An elastic collision is one during which no K.E. is lost.   This means that inelastic collisions are those with energy losses during collision.  Of course, a 100% elastic collision is ideal because during any collision, there is always some energy loss in the form of tear and wear as well as conversion to heat.  In reality, all real collisions are inelastic and are always associated with some energy loss no matter how small.  Often when we have very elastic (very hard and bouncy) materials, we may assume perfectly elastic collisions just to simplify the solutions to problems at hand. 

We may compare the elasticity of different materials by dropping a small sphere of each from a certain height (1.00m for example), and see which one bounces higher.  If one bounces back to the same 1.00m height, it will be 100% elastic!  Of course, you know that it won't happen.  One might think that a rubber ball would bounce the highest.  In fact a hard steel ball (such as one out of a ball bearing) will bounce higher than a rubber ball does when dropped on a very smooth and horizontal solid steel floor.

Chapter 7 Test Yourself :

1)  Momentum is  (a) a scalar   (b) a vector   (c) sometimes a vector and sometimes a scalar quantity.   click here.

2) Momentum is defined as the product of (a) Force and a time interval   (b) force and mass   (c) Mass and velocity.

3) The reason momentum is a vector is that (a) mass is a vector  (b) velocity is a vector   (c) neither a, nor b.

4) If your car including you has a mass of 800-kg and is moving at (25m/s, North), the momentum of your vehicle is (a) 20,000kg m/s   (b) 20,000 kg m/s, North    (c) neither a, nor b.   click here.

Problem:  Suppose you are in outer space far from planets and stars (almost zero gravity).  If you are holding a 1.0-kg rock in your hand and your mass including your space suit is 75 kg and you throw the rock in say +x direction at a speed of 7.5m/s.  Answer the following:    click here.

5) (a) You remain at rest  (b) You move at a speed of 7.5m/s in the opposite direction  (c) You move at a speed of 0.10m/s in the opposite direction.

6) The rock's momentum  is (a)+7.5kgm/s  (b)+7.5kgm/s2  (c) +7.5kg/s.   click here.

7) Your momentum after the rock is thrown is (a) -7.5kgm/s   (b) 0   (c) 75g m/s.

8) If you are at (0,0), and your friend on the negative x-axis at (-20.0m, 0), how long will it take for you to reach him?  (a) 75s  (b) (1/75)s  (c) 200s.

9) The average force a baseball bat exerts on a baseball during a contact time of 0.025s is 400N.  The magnitude of the Impulse of the bat on the baseball is (a)425 Ns  (b)10Ns   (c)16000 Ns.   click here.

10) The impulse of force F during the time interval Δt is  (a) FΔt   (b) F /Δt   (c) FΔt2.

11) FΔt  acting on mass M is equal to (a) the change in the acceleration of M  (b) the change in mass M  (c) the change in the linear momentum of M.   click here.

12) Impulse-momentum equivalence is (a) FΔt = M(vf2 - vi2)     (b) FΔt  = M(vf - vi)     (c) F = Mv.

13) For the head-on collision of two equal size balls of masses M1 and M2 moving with velocities v1 and v2,  linear momentum conservation is (a) M1v1 = M2v2    (b) M1u1 = M2u2   (c) M1v1 + M2v2 = M1u1 + M2u2  where u1and u2 are velocities after collision.

14) A perfectly elastic collision is one during which (a) there is no potential energy change  (b) K.E. remains constant  (c) one object remains at rest.   click here.

15) If you drop a ball made of an elastic material from a height of 1.00m on a rigid floor that is also made of the same material, you may call it perfectly elastic if it bounces back to a height of  (a) 0.500m    (b) 0.950m   (c) 1.00m again.

16) A perfectly elastic material is (a) ideal and cannot really be made  (b) real and easy to make  (c) called flubber.

17) In perfectly elastic collisions, (a) K.E. is conserved   (b) K.E. does not change   (c) K.E. before collision is equal to K.E. after collision  (d) a, b, & c, are the same thing and correct.   click here.

18) When a bullet hits a chunk of wood and gets embedded in it, since part of the bullet's K.E. is consumed for deformation and penetration into the wood, we may say that the collision is (a) inelastic  (b) elastic  (c) elastic but with some energy loss.

Problem:  A 0.0500-kg bullet is fired at a muzzle speed of 400m/s to the right into a 3.950-kg chunk of wood hanging from a tree via a long cord.  After collision, the wood-bullet combination gains velocity u and swings.  Answer the following questions:

19) The initial K.E. of the bullet before collision is (a) 8000J   (b) 16000J   (c) 4000J.   click here.

20) The initial K.E. of the still wood before collision is (a) 3.950J   (b) 0    (c) 400J.

21) The conservation of momentum before and after collision may be written as Mbvb + Mwvw = (Mb +Mw)u.   (a) True  (b)False    click here.

22)  The wood-bullet velocity, u, after collision is (a) 10.0m/s    (b) 5.00 m/s   (c) 0.

23) The K.E. of the wood-bullet combo, after collision is (a) 50.0J   (b) 250J   (c) 400J.

24) The change in K.E. in this collision is (a) -3950J    (b) 3900J   (c) 0.   click here.

25) This collision is (a) highly elastic   (b) highly inelastic   (c) perfectly inelastic.

Problems:  (g = 9.81m/s2 in the following problems).

1) A rifle attached to a plank of wood is placed on a horizontal long table on a track.  The rifle barrel is parallel to the table.  The coefficient of friction between the plank and the table is 0.450.  When the rifle is fired, the bullet goes to the left and the rifle-plank combo slides to the right.  The rifle-plank combo comes to stop after sliding a distance of 2.40m.  If the mass of the bullet is 69.0 grams and that of the rifle-plank combo excluding the bullet 5.30kg, find (a) the initial speed of the rifle-bullet combo just after firing, and (b) the muzzle speed (the initial speed) of the bullet just after firing.

2) A 40.0-gram rubber ball released from a height of 1.41m above a perfectly horizontal concrete floor bounces back to a height of 1.13m.  Calculate (a) its velocity just before collision, (b) its velocity just after collision, and (c) the loss in its kinetic energy.

3) A small steel ball is dropped from a height of 1.00m onto a perfectly horizontal steel floor.  If the change in the kinetic energy during collision is 5.0%, calculate the height that the ball reaches after bouncing off the floor.

4) In a collision, an 8.00-ton train car traveling at a velocity of (1.20m/s, North) interlocks with an empty train car that has a mass of 2.00 tons that is initially at rest. Calculate (a) the velocity of the interlocked cars just after collision, and (b) the change in the K.E..

5) On a horizontal surface, solid ball A (MA = 0.200kg) traveling at (vA = + 4.00m/s) makes a head-on collision with ball B (MB = 0.200kg) that is initially at rest (vB = 0).    Let the after collision velocities be uA and uB, and write (a) the momentum balance equation, (b) the energy balance equation, both in terms of uA and uB.  Solve the two equations to find the unknowns uAand uB.  Assume the collision is perfectly elastic that means there is no loss in the kinetic energy of the balls.

6) A 48-gram tennis ball traveling to the right at 25m/s is hit by a racket that exerts a leftward force of 120N on the ball for 0.030s.  Find the final velocity of the tennis ball.

7) An 80.0-gram baseball traveling horizontally to the left at 35m/s gets hit by a bat that exerts a rightward force of 320N on the ball for a short time.  The ball returns horizontally at a speed of 65m/s.  Find the contact time between the bat and the ball.

Answers: 1) 4.60m/s, 353m/s    2) -5.26m/s,  +4.71m/s,  -0.110J    3) 95cm   

4) 0.96m/s, North;  -1150J    5) uA= 0,  uB= +4.00m/s    6) -50m/s    7) 0.025s