Linear Momentum

Assume your file cabinet is sitting on a cart with wheels.  It is sitting in the middle of a room with a smooth floor.  You want to move it against the wall.  You give it a push.  It takes off, and before you know it, it slams into the wall.  It is hard to stop it, because it has linear momentum.

Linear momentum is a measure of an object's translational motion.  The linear momentum p of an object is defined as the product of the object's mass m times its velocity v.

p = mv.

Linear momentum is a vector.  Its direction is the direction of the velocity.  The Cartesian components of p are

px = mvx,  py = mvy,  pz = mvz.

If an object's velocity is changing, its linear momentum is changing.  We have

dp/dt = d(mv)/dt.

If the mass of the object is constant then

dp/dt = mdv/dt = ma.

We write

dp/dt = F.

This is a more general statement of Newton's second law which also holds for objects whose mass is not constant.

If an object receives an impulse its momentum changes.  We may write

dp = Fdt.

Therefore

.

If the force acting on the object is constant, then

Dp = FDt.

The integral of force over time is called the impulse I of the force.  We have shown that the impulse I is equal to the change in momentum Dp.  You give an object an impulse, by letting a force act on it for a time interval Dt.

I = Dp = FavgDt

Note the difference:

Work: W = F×d  (scalar)
Impulse: I = Dp = FDt   (vector)

Problems:

A car is stopped for a traffic signal.  When the light turns green, the car accelerates, increasing its speed from 0 to 5.2 m/s in 0.832s.  What linear impulse and average force does a 70 kg passenger in the car experience?
Solution:

Assume the car accelerates in the x-direction.  The final momentum of the passenger is pxf = 70kg5.2m/s = 364kgm/s.  The initial momentum is zero. 
Dpx = pxf - pxi = 364kgm/s = FavgDt.  Favg = (364kgm/s)/0.832s = 437.5N in the x-direction.

A 3kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60o with the surface.  It bounces off with the same speed and angle.  If the ball is in contact with the wall for 0.2s, what is the average force exerted on the ball by the wall?

Solution:

The balls initial momentum is
pi = pxii + pyij = (3kg10m/s)sin60oi+(3kg10m/s)cos60oj.

Its final momentum is
pf = pxfi+pyfj = -(3kg10m/s)sin60oi+(3kg10m/s)cos60oj.

Dp = pf-pi = -2(30kgm/s)sin60oi = -(51.96kgm/s)i.
Dp = Favg Dt.
Favg = -(51.96kgm/s)i/(.2s) = -259.8Ni.

Does a large force always produce a larger impulse on a body than a smaller force does?  Explain!
Solution:

The impulse is the integral of force over time, I = Dp = FavgDt.  A small force acting over a long time can produce a larger impulse than a large force acting over a short time.

Exercise  (You can earn up to 5 points extra credit by completing this exercise.)

Conservation of momentum

Consider two interacting objects.  If object 1 pushes on object 2 with a force F = 10N for 2s to the right, then the momentum of object 2 changes by 20Ns = 20kgm/s to the right.  By Newton's third law object 2 pushes on object 1 with a force F = 10N for 2s to the left.  The momentum of object 1 changes by 20Ns = 20kgm/s to the left.  The total momentum of both objects does not change.  For this reason we say that the total momentum of the objects is conserved.

Newton's third law implies that the total momentum of a system of interacting objects that are not acted on by outside forces is conserved.

The total momentum in the universe is conserved.  The momentum of a single object, however, changes when a net force acts on the object for a finite time interval.  Conversely, if no net force acts on an object, its momentum is constant.  For a system of objects, a component of the momentum along a chosen direction is constant, if no net outside force with a component in this chosen direction acts on the system.

Problems:

A 0.1kg ball is thrown straight up into the air with an initial speed of 15m/s.  Find the momentum of the ball
(a) at its maximum height and
(b) half way up to its maximum height.
Solution:
(a) At the ball's maximum height its velocity is zero, and therefore its momentum is zero.
(b) The ball's total energy E = K + U is constant.  
Initially U = 0 and K = (1/2)mv2 = 0.05kg(15m/s)2 = 11.25J.  
At its maximum height K = 0 and U = mgh = 11.25J.
Halfway up to its maximum height K = U = (1/2)11.25J.  
Therefore v2 = 2K/m = 11.25J/.1kg = 112.5(m/s)2, 
v = 10.6m/s, p = 0.1kg10.6m/s = 1.06kgm/s.
Two blocks of mass M and 3M are placed on a horizontal frictionless surface.  A light spring is attached to one of them, and the blocks are pushed together with the spring between them.  A cord holding them together is burned, after which the block of mass 3M moves to the right with a speed of 2m/s.
(a) What is the speed of the block of mass M?
(b) Find the original elastic potential energy in the spring if M = 0.35kg.

Solution:
(a)  The total initial momentum of the system is zero.  No outside force act on the system in the horizontal direction, so the final horizontal momentum component of the system is also zero.

p3M + pM = 0.

p3M = 3M2m/s, pM =- 3M2m/s = -6Mm/s.  vM = -6m/s.  The block of mass M moves towards the left with a speed of 6m/s.

(b) The total energy E = K + U of the system is conserved.  After the string has been cut and the spring has relaxed, E = K = (1/2)3M(2m/s)2+(1/2)M(6m/s)2 = 8.4J.  Initially E = U = 8.4J, the total energy is the elastic energy stored in the spring.

Please read:  Analysis of an accident

Collisions

In collisions between two isolated objects Newton's third law implies that momentum is always conserved.  Collisions in which the kinetic energy is also conserved, i.e. in which the kinetic energy just after the collision equals the kinetic energy just before the collision, are called elastic collision.  In these collisions no ordered energy is converted into thermal energy.  Collisions in which the kinetic energy is not conserved, i.e. in which some ordered energy is converted into internal energy, are called inelastic collisions.  If the two objects stick together after the collision and move with a common velocity vf, then the collision is said to be perfectly inelastic.

Note: In collisions between two isolated objects momentum is always conserved.

We always have
m1v1i + m2v2i = m1v1f + m2v2f.

Kinetic energy is only conserved in elastic collisions.  Only for elastic collisions do we also have
(1/2)m1v1i2 + (1/2)m2v2i2 = (1/2)m1v1f2 + (1/2)m2v2f2.

Problems:

A 10g bullet is stopped in a block of wood (m=5kg).  The speed of the bullet-wood combination immediately after the collision is 0.6m/s.  What was the original speed of the bullet?
Solution:

This is a perfectly inelastic collision.  The two objects stick together after the collision and move with a common velocity vf.  Assume the motion is along the x-direction. We then have

m1v1 + m2v2 = (m1 + m2)vf.

m1 is the mass of the bullet, v1 its initial velocity, m2 is the mass of the block, v2 its initial velocity.  Initially the block is at rest, v2 = 0.  Therefore

0.01kgv1 = 5.01kg0.6m/s. v1 = 300.6m/s.

A neutron in a reactor makes an elastic head-on collision with the nucleus of a carbon atom initially at rest.
(a) What fraction of the neutron's kinetic energy is transferred to the carbon nucleus?
(b) If the initial kinetic energy of the neutron is 1.6´10-13J, find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision.
(The mass of the carbon nucleus is about 12 times the mass of the neutron.)
Solution:
(a) Initially the carbon nucleus is at rest.   Let particle 1 be the neutron and particle 2 be the carbon nucleus.   Assume the motion is along the x-direction.  We have

(i) m1v1i  = m1v1f + m2v2f.

(ii) (1/2)m1v1i2  = (1/2)m1v1f2 + (1/2)m2v2f2.

We can solve this system of two equations for the ratio v2f/v1i. We obtain

v2f2 = (m1/m2)(v1i2 - v1f2)   from (ii),

v1f = v1i - (m2/m1)v2f    from (i).

We can therefore write

v1f2 = v1i2 + (m2/m1)2v2f2 - 2(m2/m1)v1iv2f.

v2f2 = (m1/m2)(2(m2/m1)v1iv2f  - (m2/m1)2v2f2) = (2v1iv2f  - (m2/m1)v2f2).

(1+(m2/m1))v2f = 2v1i.

v2f = 2m1v1i/(m1 + m2).

v2f/v1i = 2m1/(m1 + m2) = 2/13 = 0.15

The fraction of the kinetic energy transferred to the carbon nucleus is (m2/m1)(v2f/v1i)2 = 12(0.15)2 = 0.284.

(b) The initial kinetic energy of the of the neutron is 160fJ and the final kinetic energy of the carbon nucleus is 45.4fJ. (1femtoJoule = 1fJ = 10-15J.)  The final kinetic energy of the neutron is (160 - 45.4)fJ.
A 90kg fullback running east with a speed of 5m/s is tackled by a 95kg opponent running north with a speed of 3m/s.  If the collision is perfectly inelastic, calculate the speed and the direction of the players just after the tackle.
Solution:

The initial momentum of player 1 is p1 = 90kg5m/si = 450kgm/si.
The initial momentum of player 2 is p2 = 95kg3m/sj = 285kgm/sj.
Momentum is conserved, the final momentum p of both players is p = p1+p2.

p = (m1 + m2)v.
v = 2.432m/si+1.54m/sj.
v2 = 8.29(m/s)2, v = 2.88m/s.
The speed of the players after the collision is 2.88m/s.
tanq = py/px = 285/450 = 0.63, q = 32.34o.
Their direction of travel makes an angle q = 32.34o with the x-axis. (The x-axis is pointing east.)

The mass of the blue puck is 20% greater than the mass of the green one.  Before colliding, the pucks approach each other with equal and opposite momenta, and the green puck has an initial speed of 10m/s.  Find the speed of the pucks after the collision, if half the kinetic energy is lost during the collision.

Solution:

The total momentum of the two pucks is zero before the collision and after the collision.

Let particle 1 be the green puck and particle 2 be the blue puck.  Before and after the collision the ratio of the speeds is v2/v1 = m1/m2 = 1/1.2.

The final kinetic energy of the system equals 1/2 time its initial kinetic energy.
(1/2)m1v1i2 + (1/2)m2v2i2 = 2((1/2)m1v1f2 + (1/2)m2v2f2).
m1v1i2 + m2v2i2 = 2(m1v1f2 + m2v2f2).
m1v1i2 + 1.2m1(v1i/1.2)2 = 2(m1v1f2 + 1.2m1(v1f/1.2)2).
m1(v1i2 + v1i2/1.2) = 2m1(v1f2 + v1f2/1.2).
v1i2 = 2v1f2, v1f = 0.707v1i.
v1f = 7.07m/s.  v2f = 5.89m/s.

If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision?  Is it possible for one to be at rest after the collision?  Explain!
Solution:

In collisions between two objects momentum is conserved.  Since the initial momentum is not zero, the final momentum is not zero.  Both objects cannot be at rest.
It is possible for one of the objects to be at rest after the collision.  For example, if the masses of the two objects are equal, then after a head-on elastic collision the object initially at rest is moving and the object initially moving is at rest.

Link:  

1-d collisions

More links:

Momentum and Collisions

Balls hitting balls

Assume two balls of equal mass, made from the same material, approach each other head on.  Both balls have the same speed v.  They approach each other with relative speed 2v.  As the balls collide, each ball exerts a force on the other.  The forces are equal in magnitude but have opposite directions.  The balls distort like spherical springs, and the same amount of energy is stored in each ball as elastic potential energy.  It will be reconverted into kinetic energy.  The force with which ball 1 pushes on ball 2 first decelerates ball 2 to a stop and then accelerates it into a direction opposite its initial velocity.  The force with which ball 2 pushes on ball 1 decelerates ball one to a stop and then accelerates it into a direction opposite its initial velocity.  We expect the two balls to fly apart with equal speeds in opposite directions.  If the coefficient of restitution of the two balls is 1, then their speed will not have changed.  The total momentum of the two balls is conserved.

Now suppose you are looking at this collision from a different reference frame.  Suppose as ball 1 approaches ball 2, you are sitting in a chair that rolls with the same velocity right along side ball 1.  With respect to you, ball 1 does not move.  Ball 2 approaches ball 1 with speed 2v, the relative speed of the balls. After the collision ball 2 has the same velocity as you do. So ball 2 now does not move with respect  to you, but ball 1 now moves backward with speed 2v.  In your reference frame ball 2 hits a stationary target. It comes to rest and ball 1 leaves the collision with a velocity equal to the incoming velocity of ball 1.  Momentum is conserved.

Both reference frames are valid reference frames in which to describe the collision.  In any reference frame which is not accelerating, i.e. in any inertial frame, Newton's laws are valid.

People in different reference frames see different things.  They do not agree on the velocity, momentum, or kinetic energy of objects.  They will, however, always agree on the relative velocity of two objects, and they will always be able to use Newton's laws in their reference frame to explain what they are observing.

Bats and baseballs

Assume a baseball hits a stationary bat.  If the bat is nailed to the wall of a house and cannot move at all, then the ball will just rebound the same way it rebounds from a hard floor.  If the bat is held in the hand of the batter, then the force the ball exerts on the bat will accelerate the bat backwards, and some collision energy will be transferred to the bat and will not appear as rebound energy of the ball.

If the batter and the bat are very heavy, they receive little of the collision energy, and the ball rebounds with outgoing speed equal to the coefficient of restitution times the incoming speed.

If the batter is swinging the bat forward as the ball hits it, the ball's outgoing speed will be much higher. Assume the bat and the ball each are moving with speed 100 km/h in opposite direction. Their relative speed is 200 km/h. A reference frame in which the bat is stationary is moving with 100 km/h speed with the bat. In this reference frame the ball approaches with 200 km/h and rebounds with speed v = coefficient of restitution times 200 km/h in the forward direction.  But this reference frame is moving itself with speed 100 km/h in the forward direction. With respect to the ground the ball is therefore moving with speed vground = 100km/h + v = 100km/h + coefficient of restitution * 200 km/h.

Example:
A baseball heads toward home plate at 100 km/h.
The bat heads toward the pitcher at 100 km/h.
The relative speed between ball and bat as they are approaching each other is 200 km/h.

Assume the baseballs coefficient of restitution is 0.55.  Just after the collision the relative speed between ball and bat is 0.55*200 km/h = 110 km/h.
The bat still heads towards the pitcher at approximately 100 km/h.

The ball moves relative to the bat with a speed of 110 km/h towards the pitcher.

Relative towards home plate and the pitcher, the ball's speed is 210 km/h.  The ball heads toward pitcher at 210 km/h
 

 Animation: Ball hitting a stationary bat

Animation: Ball hitting a moving bat

Links to other Web Materials:

Momentum and its conservation

Linear momentum and collisions

Momentum conservation in nuclear physic

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