Linear Momentum

Unit 5

Lesson 1 : Linear Momentum and Its Conservation

F21 + F12 = 0(Newton’s Third Law)

m1a1 + m2a2 = 0(Newton’s Second Law)

m1

dv1

dtm2

dv2

dt+ = 0

d(m1v1)

dt

d(m2v2)

dt+ = 0

(m1v1 + m2v2) = 0d

dt

sum of linear momentum is constant

p = mv (linear momentum)

Linear momentum is a vector quantity whose direction the same as the

direction of v. Its SI unit is kg . m/s.

F = ma = mdv

dt

F =d(mv)

dt

dp=

dt

The time rate of change of the linear momentum of a particle is equal to the

net force acting on the particle.

(This is the form in which Newton presented his second law.)

Conservation of Linear Momentum

Whenever two or more particles in an isolated system interact, the total momentum of the

system remains constant.

The total momentum of an isolated system at all times equals its initial momentum.

Total p before = Total p after

A 60. kg archer stands at rest on frictionless ice and fires a 0.50 kg arrow horizontally at

50. m/s.

Example 1

a) With what velocity does the archer move across the ice after firing the arrow.

b) What if the arrow were shot in a direction that makes an angle with the

horizontal ? How will this change the recoil velocity of the archer ?

Two blocks of masses 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 initially holding the

blocks together is burned, and the 3M block moves to the right with a

speed of 2.00 m/s.

Example 2

a) What is the speed of the block of mass M ?

b) Find the original elastic potential energy in the spring if M = 0.350 kg.

A swing seat of mass M is connected to a fixed point P by a massless cord of length L. A child also of mass M sits on the

seat and begins to swing with zero velocity at a position at which the cord makes a 60o angle with the vertical as shown

in Figure I. The swing continues down until the cord is exactly vertical at which time the child jumps off in a

horizontal direction.

Example 3 : AP 1981 #2

The swing continues in the same direction until the cord makes a 45o angle with the vertical as shown in Figure II; at that point it begins to swing in the reverse direction. With what velocity relative to the ground did the child leave the swing ? (cos 45o = sin 45o = 2

2

, sin 30o = cos 60o = ½,

cos 30o = sin 60o = 3

2)

Lesson 2 : Impulse and Momentum

F =d(mv)

dt

dp=

dt

dp = F dt

Integrating F with respect to t,

p = pf – pi = F dttf

ti

Impulse (I)

F dt

tf

ti

I =

Impulse has a magnitude equal to the area under the force-time

graph.

Impulse is a vector quantity with the same direction as the direction of the change in momentum.

Because the force imparting an impulse can generally vary in time, we can express impulse as

I = Ft

Impulse – Momentum Theorem

p = pf – pi = F dttf

ti

The impulse of the force F acting on a particle equals the change in the

momentum of the particle.

I = p

In a particular crash test, a car of mass

1500 kg collides with a wall, as shown. The

initial and final velocities of the car

are vi = -15.0i m/s and vf = 2.60i m/s, respectively.

^

^

Example 1

a) If the collision lasts for 0.150 s, find the impulse caused by the collision and the average force exerted on the car.

b) What if the car did not rebound from the wall ? Suppose the final velocity of the car is zero and the time interval of the collision remains 0.150 s. Would this represent a larger or a

smaller force by the wall on the car ?

A 3.00 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0o with the

surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.200 s, what is the average force exerted

by the wall on the ball ?

Example 2

An estimated force-time curve for a baseball struck by a bat is shown above.

From this curve, determine

Example 3

a) the impulse delivered to the ball

b) the average force exerted on the ball

c) the peak force exerted on the ball

Lesson 3 : Collisions in One-Dimension

Elastic Collisions

The total KE (as well as total momentum) of the

system is the same before and after the collision.

Inelastic Collisions

The total KE of the system is not the same before and

after the collision (even though the momentum of the system is conserved).

KE is conserved KE is not conserved

“perfectly inelastic” “inelastic”

m1v1i + m2v2i = (m1 + m2)vf

Perfectly Inelastic Collisions

when the colliding objects stick together

Elastic Collisions

m1v1i + m2v2i = m1v1f + m2v2f

½ m1v1i2 + ½ m2v2i

2 = ½ m1v1f2 + ½ m2v2f

2

Example 1

An 1800 kg car stopped at a traffic light is struck from the rear by a 900 kg car, and the two become entangled, moving along the same path as that of

the originally moving car.

a) If the smaller car were moving at 20.0 m/s before the collision, what is the velocity of the

entangled cars after the collision ?

b) Suppose we reverse the masses of the cars – a stationary 900 kg car is struck by a moving 1800 kg car. Is the final speed the same as

before ?

The ballistic pendulum is an apparatus used to measure the speed of a fast-moving projectile, such as a bullet. A bullet of mass m1 is fired into a large block of wood of mass m2 suspended from some light wires. The bullet

embeds in the block, and the entire system swings through a height h. How can we determine the speed of

the bullet from a measurement of h ?

Example 2

A block of mass m1 = 1.60 kg initially moving to the right with a speed of 4.00 m/s on a frictionless

horizontal track collides with a spring attached to a second block of mass m2 = 2.10 kg initially moving

to the left with a speed of 2.50 m/s. The spring constant is 600 N/m.

Example 3

a) Find the velocities of the two blocks after the collision.

b) During the collision, at the instant block 1 is moving to the right with a velocity of +3.00 m/s, determine the velocity of block 2.

A 5 kg ball initially at rest at the edge of a 2 m long, 1.2 m high frictionless table, as shown

above. A hard plastic cube of mass 0.5 kg slides across the table at a speed of 26 m/s and strikes the ball, causing the ball to leave the table in the

direction in which the cube was moving.

Example 4 : AP 1995 #1

The figure below shows a graph of the force exerted on the ball by the cube as a function of time.

a) Determine the total impulse given to the ball.

b) Determine the horizontal velocity of the ball immediately after the collision.

c) Determine the following for the cube immediately after the collision.

i. Its speed

ii. Its direction of travel (right or left), if moving

d) Determine the kinetic energy dissipated in the collision.

e) Determine the distance between the two points of impact of the objects with the floor.

A 2 kg block and an 8 kg block are both attached to an ideal spring (for which k = 200 N/m) and both are initially at rest on a horizontal frictionless surface, as shown in

the diagram above.

In an initial experiment, a 100 g (0.1 kg) ball of clay is thrown at the 2 kg block. The clay is moving horizontally

with speed v when it hits and sticks to the block. The 8 kg block is held still by a removable stop. As a result, the

spring compresses a maximum distance of 0.4 m.

Example 5 : AP 1994 #1

a) Calculate the energy stored in the spring at maximum compression.

b) Calculate the speed of the clay ball and 2 kg block immediately after the clay sticks to the block but before the spring compresses significantly.

c) Calculate the initial speed v of the clay.

In a second experiment, an identical ball of clay is thrown at another identical 2 kg block, but this time the stop is

removed so that the 8 kg block is free to move.

d) State whether the maximum compression of the spring will be greater than, equal to, or less than 0.4 m.

Explain briefly.

e) State the principle or principles that can be used to calculate the velocity of the 8 kg block at the instant that the spring regains its original length. Write the appropriate equation(s) and show the numerical substitutions, but do not solve for the velocity.

Lesson 4 : Two-Dimensional Collisions

m1v1i = m1v1f cos + m2v2f cos

0 = m1v1f sin - m2v2f sin

Since this is an elastic collision, KE is also conserved.*

* If this were an inelastic collision, KE would not be conserved, and

this equation does not apply.

½ m1v1i2 = ½ m1v1f

2 + ½ m2v2f2

A 1500 kg car traveling east with a speed of 25.0 m/s

collides at an intersection with a 2500 kg van traveling north at a speed of 20.0 m/s.

Find the direction and magnitude of the velocity of

the wreckage after the collision, assuming that the vehicles undergo a perfectly

inelastic collision (that is, they stick together).

Example 1

In a game of billiards, a player wishes to sink a target ball in the corner pocket. If the angle to the corner

pocket is 35o, at what angle q is the cue ball deflected ? Assume that friction and rotational motion are

unimportant and that the collision is elastic. Also assume that all billiard balls have the same mass m.

Example 2

Lesson 5 : The Center of Mass

Center of mass (CM) is the average position of the system’s mass

Center of mass is located on the line joining the two particles and is closer to the particle

having the larger mass.

System rotates clockwise when F is applied between the less massive particle and CM.

System rotates counterclockwise when F is applied between the more massive particle and CM.

System moves in the direction of F without rotating when F is

applied at CM.

How to Locate the Center of Mass

xCM =m1x1 + m2x2

m1 + m2

Center of mass for a system of many particles

xCM =m1x1 + m2x2 + m3x3 + ……

m1 + m2 + m3 + ……

xCM =mixi

M

Y coordinates of CM

yCM =miyi

M

Z coordinates of CM

zCM =mizi

M

Using a Position Vector (r) to locate CM of a system of particles

rCM = xCM i + yCM j + zCM k^ ^ ^

Using a Position Vector (r) to locate CM of an extended object

xCM =ximi

Mlim

mi 0

(likewise for yCM and zCM)

=1

Mr dm

rCM

=1

Mx dm

xCM

Suspend the object first from points A and C.

Locating CM of an irregularly shaped object

CM is where lines AB and CD intersect

If wrench is hung freely from any point, the vertical line

through this point must pass through CM.

A baseball bat is cut at the location of its center of mass. The piece with smaller mass is

Example 1

a) the piece on the right.

b) the piece on the left.

c) both have same mass.

d) impossible to determine.

A system of three particles located as shown above. Find the center of mass of the system.

Example 2

a) Show that the center of mass of a rod of mass M and length L lies midway

between its ends, assuming the rod has a uniform mass per unit length.

Example 3

b) Suppose a rod is nonuniform such that its mass per unit length varies

linearly with x according to the expression = x, where is a constant. Find the x coordinate of the center of mass as a fraction of L.

Lesson 6 : Motion of a System of Particles

Velocity of CM

vCM =drCM

dt

Acceleration of CM

aCM =dvCM

dt

Total momentum of a system of particles

MvCM = mivi = pi = ptot

(i is the ith particle)

Newton’s Second Law for a system of particles

Fext = MaCM

total mass of system

CM of a system of particles of combined mass M moves like an equivalent particle of mass M

would move under the influence of the net external force on the system.

Suppose you tranquilize a polar bear on a smooth glacier as part of a research

effort. How might you estimate the bear’s mass using a measuring tape, a rope, and

knowledge of your own mass ?

Example 1

Example 2

Consider a system of two particles in the xy plane :

m1 = 2.00 kg is at r1 = (1.00 i + 2.00 j) m and has a velocity of (3.00 i + 0.500 j) m/s

m2 = 3.00 kg is at r2 = (-4.00 i - 3.00 j) m and has a velocity of (3.00 i – 2.00 j) m/s

^ ^

^ ^

^ ^

^ ^

a) Plot these particles on a grid or graph paper. Draw their position vectors and show their velocities.

b) Find the CM of the system and mark it on the grid.

c) Determine the velocity of the CM and also show it on the diagram.

d) What is the total linear momentum of the system ?

Example 3

A rocket is fired vertically upward. At the instant it reaches an altitude of 1000 m and a speed of 300 m/s, it explodes into three

fragments having equal mass. One fragment continues to move upward with a speed of 450 m/s following the explosion. The second fragment has a speed of 240

m/s and is moving east right after the explosion. What is the velocity of the third

fragment right after the explosion ?