M OMENTUM IN 2 D IMENSIONS. 4.1 N EWTON ' S SECOND LAW AND MOMENTUM Newton's second law states that the acceleration produced by a force on an object.

MOMENTUM IN 2 DIMENSIONS

4.1 NEWTON'S SECOND LAW AND MOMENTUM Newton's second law states that the acceleration

produced by a force on an object is directly proportional to the net force and inversely proportional to the mass of the object

This can be expressed in the form of the equation: F = ma

Acceleration can be expressed in terms of initial and final velocity, i.e.

Substituting this expression for a in the equation for Newton's second law gives:

MOMENTUM The product of mass and velocity is called

momentum. Momentum is given the symbol p and its unit

is kg.ms-1. The momentum of an object is a vector in the same direction as the velocity of the object.

The plural of momentum is momenta.

Force can therefore be expressed in terms of change in momentum:

PARTICLES REBOUNDING FROM A SURFACE

When a particle collides with a surface, it changes direction, i.e. it accelerates.

As the particle hits the surface it exerts a force on the surface.

From Newton's third law, the surface exerts an equal and opposite force on the particle, and it is this force which causes the particle to change direction.

Since a force is acting there must also be a change in momentum.

COLLISION AT RIGHT ANGLES TO A SURFACE, REBOUNDING WITH NO CHANGE IN SPEED

A particle of mass m collides at right angles with a surface. The particle rebounds at right angles at the same speed, i.e. vf = -vi.

i.e.. the average acceleration of the particle is in the direction of vf, which is at right angles away from the surface.

FORCE OF THE WALL ON THE PARTICLE IS GIVEN BY:

From Newton's second law the force of the wall on the particle is given by:

The force, acceleration and change in momentum are all in the same direction, i.e. atright angles away from the wall.

4.3.2 EXAMPLE 4.1

A ball of mass 210 g travelling at 25 ms-' collides with a wall at right angles with no loss of speed. The ball is in contact with the wall for 0.050 s. Find: (a) the change in velocity of the ball (b) the acceleration of the ball (c) the change in momentum of the ball (d) the force exerted by the wall on the ball (e) the force exerted by the ball on the wall.

COLLISION AT AN ANGLE TO THE SURFACE AND REBOUNDING WITH NO CHANGE IN SPEED

Find two components for each of - vi and vf One component is parallel to the wall, the other component is perpendicular to the wall.

EXAMPLE 4.2 A ball of mass 50 g meets a wall at an angle of 50°,

with a speed of 5.0 ms-1. It rebounds with the same speed after being in contact with the wall for 0.10 seconds. Find: (a) the change in velocity of the ball (b) the average acceleration of the ball (c) the change in momentum of the ball (d) the force exerted by the wall on the ball (e) the force exerted by the ball on the wall.

EX 4.3

A cricket ball of mass 150 g is travelling North at 20 ms-l. After being hit by a cricket bat, the ball travels East at 25 ms-l. The ball is in contact with the bat for 0.2 sec. Find the force exerted by the bat on the ball

ELASTIC AND INELASTIC COLLISIONS

Interactions such as collisions may be termed as either elastic or inelastic depending on the energy transformations that take place.

An elastic collision is one in which the kinetic energy is conserved (eg. none of the KE is converted to other forms of energy).

An inelastic interaction (or non elastic interaction) is an interaction in which kinetic energy is not conserved (eg. some of the kinetic energy is converted to other forms of energy)

4.4 CONSERVATION OF MOMENTUM

A system in which no external forces are acting is said to be isolated.

When the cricket ball and basketball collide, they exert forces on each other, which are called internal forces because they are within the system and are only caused by the collision between the two objects.

From Newton's Third Law, the force exerted by the cricket ball on the basketball must be equal and opposite to the force exerted by the basketball on the cricket ball.

4.4.1 LAW OF CONSERVATION OF MOMENTUM

LAW OF CONSERVATION OF MOMENTUM From Newton's third law:

Δt is the time during which the two balls are in contact and so it must have the same value for both. Therefore:

This means that the overall change in momentum of the two objects in the system is equal to zero.

The total momentum of any number of objects remains unchanged, regardless of the number of collisions between the objects, provided no external forces are acting.

4.4.2 CALCULATIONS

Always draw before and after diagrams for questions involving conservation of momentum.

Make sure that your vector diagrams are adding momentum not velocity. Velocity is not necessarily conserved in isolated collisions.

EXAMPLE 4.4 -TWO OBJECTS TRAVELLING IN A STRAIGHT LINE COLLIDE AND STICK TOGETHER

A gun fires a bullet of mass mb = 55 g, travelling with a velocity of vbi = 250ms-1 W, into a stationary block of wood of mass mb = 2.0 kg. The bullet remains embedded in the wood. Find: (a) the total initial momentum of the system (b) the final velocity of the wood and bullet.

EXAMPLE 4.5 -TWO OBJECTS TRAVELLING IN A STRAIGHT LINE COLLIDE BUT DO NOT STICK TOGETHER

A train engine on straight railway tracks has a mass, m, = 1.2 x 104 kg, and is travelling with a velocity, vei = 15 kmph. It collides with a carriage of mass, m, = 6.0 x 103 kg, travelling in the same direction with a velocity of vci = 5.0 kmph. The carriage moves off after the collision with a velocity of vcf = 20 kmph. What is the final velocity of the engine?

Let the initial direction of the engine be positive.

Find the initial momenta of the engine and the carriage.

Use conservation of momentum to find the final momentum of the engine.

Find the final velocity of the engine from its final momentum.

i.e.. the final velocity of the engine is 2.1 ms-l in its original direction.

EXAMPLE 4.6 -TWO OBJECTS TRAVELLING AT RIGHT ANGLES COLLIDE AND DO NOT STICK TOGETHER

A boy on roller blades with mass, mb = 55 kg, travelling North with a speed of vbi = 4ms-1, collides with a girl also on roller blades with mass, mg = 45 kg, travelling West with a speed of vgi = 3ms-1.

After the collision the girl finds herself travelling North with a speed of vgf = 2ms-1.

Find: (a) the total initial momentum (b) the velocity of the boy after the collision

(a) Find the total initial momentum by finding the vector sum of the initial momenta of the boy and girlFind the initial momentum of the boy, pbi, and the girl, pgi.

(b) Use conservation of momentum to find the boy's final velocity

Total initial momentum = total final momentum.

EXAMPLE 4.7 -TWO OBJECTS COLLIDE, NOT AT RIGHT ANGLES, AND DO NOT STICK TOGETHER

Two balls are rolling on a frictionless surface when they collide as shown in the diagram.

ANSWER

Let the initial direction of ball 1 be North. This is an acceptable way of determining direction as long as it is clearly stated at the beginning of the question.

Find the initial momenta of Ball 1, p1i, and Ball 2, p2i.

Find the total initial momentum, pi, of Balls 1 and 2 by finding the vector sum of their initial momenta.Use the cosine rule to find side c which is pi.

Use the sine rule to find angle B which is .

i.e. the total initial momentum is 0.50 sN N 16° W.

4.5 MULTI-IMAGE PHOTOGRAPHS OF COLLISIONS ON AN AIR-TABLE An air-table is a flat

surface with many small holes through which air is blown.

This forms a cushion which reduces friction almost to zero.

Pucks are small, round, flat objects which float on the air cushion and move freely around the air-table.

The end result is similar to that shown right.

Flash rate = 5 per secondScale = 1:10

4.5 MULTI-IMAGE PHOTOGRAPHS OF COLLISIONS ON AN AIR-TABLE The distance between

successive images is a measure of the speed, (since the time interval is constant).

The direction being given by the photograph.

Thus joining two successive images represents the magnitude and direction of the velocity vector.

Flash rate = 5 per secondScale = 1:10

4.5 MULTI-IMAGE PHOTOGRAPHS OF COLLISIONS ON AN AIR-TABLE By using the scale and

flash rate used in the photograph we can obtain the actual speed of the pucks

Distance, s = distance between images adjusted by the scale (1 cm rep. 0.1 m) to give real distance.

Time, (t )= time between flashes = 1/5 s = 0.2 s.

Flash rate = 5 per secondScale = 1:10

4.5.1 VELOCITY OF A PUCK