Chapter 7

Impulse and Linear Momentum

Linear Momentum is defined as the product of mass and linear velocity.  Pushing an object to the right,  results in a 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 both mathematically and experimentally verify that

Mb Vb   =   MrVr .       This means:      " Momentum to the left "    =     " Momentum to the right ."

The product  MV , or linear momentum is an important quantity in physics.

Momentum is a Vector:

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 speed 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.

Solution: (a)  M1V1 = (0.15 kg)(+4.0 m/s) =  + 0.60 kg m/s  ;   M2V2  = (0.35 kg)(-6.0 m/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 is served by a pitcher horizontally to the left at 17 m/s and it returns to the right at 63 m/s after getting struck by a bat.  Calculate the change in its momentum.

Solution: Recall that Δ is used to denote change and it means a final value minus its initial value; therefore, we need to calculate Δ(MV) = MVf - MVi.

Δ(MV) = M ( Vf - Vi ) = 0.120kg [ ( + 63 m /s) - ( - 17 m /s) ]  =  + 9.6 kg m/s.

Impulse:

Impulse ( I ) is the product of force and a time interval.  Mathematically impulse ( I ) 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 = F Δt  ;  I = ( - 44N)(25s) = -1100 Ns.

Note that Impulse and momentum have same units.  The unit of momentum MV is (kg m/s).

So is the unit of impulse (Ns) =  (kg m/s2)(s)  =   kg m/s.

Equivalence of Impulse and Linear Momentum:

It is easy to show that the impulse of force F during time Δt on mass M is equal to the change in the linear momentum of mass M, simply,

FΔt = Δ(MV)    ;        F Δt  =  M Vf  -  M Vi        or,

Impulse = Change in Momentum.

Proof:   Starting with Newton's 2nd law (for a single force):  (Make sure to write all down as you proceed using horizontal fraction bars).

F = Ma  , replacing  a  by   Δv/Δt , and multiplying thru by Δt, results in:

F = M Δv/Δt   ; FΔt = M Δv   ; FΔt = M (vf - vi)   ; FΔt  = Mvf - M vi .

and the equivalence of  Impulse and change in linear momentum is verified.

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 15 m/s to the left and after the strike  becomes 45 m/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  =    ( + 45 m/s )  -  ( -15 m/s )  =  + 60. m/s    (The change in velocity, not the change in speed ! )

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

(c)  According to the equivalence of impulse and linear 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:

It is easy to show that when a system of particles go through collisions with each other, the total momentum remains constant.  To prove this, let us consider the head-on collision of only two balls that move toward each other along the same straight line.  Also suppose that both balls have the same size.  The following figure shows solid spheres A and B with masses M1 and M2 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:

The total momentum before collision is:   M1V1  +  M2V2       ( V1 and V2 are velocities before collision).

The total momentum after collision is    M1 u1  +  M2 u2        ( u1 and u2 are velocities after collision).

During collision, each ball acts as a wall for the other.  In fact, each ball act 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 so are the contact times.

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:  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 itself moves to the left at 2.50 m/s.  Calculate the bullet exit speed.

Solution:  Total mom. after collision must be equal to the total mom. 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 mom. after firing must also be equal to zero as well. This means that:

MbVb + MrVr = MbVb + MrVr     ;    (Note that Mr is not just the mass of rifle, it is the mass of rifle and cart).

(0.045kg)( 0 ) + (4.50kg + 1.50kg)( 0 )   =  (0.045kg)(Vb) + (4.50kg + 1.50kg)( - 2.50 m/s)    or,

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

or,                                   15.0 kg m/s  =   0.045 Vb

Vb  =  + 333 m/s    ( Of course, (+) means to the right )

Chapter 7 Test Yourself 1:

1)  Momentum is  (a) a scalar   (b) a vector    (c) sometimes a vector and sometimes a scalar.   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 (25 m/s, North), the momentum of your vehicle is (a) 20,000 kg 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 stationary  (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 momentum of the rock 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.

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

Impulse:

9) The average force a baseball bat exerts on a baseball during a contact time of 0.025s is 400N.  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) The correct form of impulse-momentum equivalence is (a)  FΔt  = M(Vf2 - Vi2)     (b) FΔt  = M(Vf - Vi)     (c) F = MV.

13) For a head-on collision of two equal size balls of masses M1 and M2 moving with velocities V1 and V2, the conservation of linear momentum 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 stationary.    click here

15) If you drop a ball made of an elastic material from a height of 1m 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) 0.5m    (b) 0.95m   (c) 1.0m 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 a perfectly elastic collision, (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, and c, mean the same thing.    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 400. m/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 a velocity V and swings.  Answer the following questions:  (All numbers are good to 3 significant figures).

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)V.   (a) True  (b)False    click here

22) From 21, the wood-bullet velocity, V, 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) Based on the results, 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%, find the maximum height the ball reaches after collision.

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. 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 uA and uB.  Assume the collision is perfectly elastic that means there is no loss in kinetic energy.

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 ball.

7) An 80.0-gram baseball traveling horizontally to the left at 35m/s gets hit by a bat that exerts a leftward 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