Momentum

Momentum is by definition the product of mass and velocity. So strictly speaking momentum is a vector quantity.

momentum = mass(kg) x velocity(ms^-1)

Hence the unit of momentum is (kg.ms^-1).

Impulse of a force

This is simply the force multiplied by the time the force acts.

We can obtain an expression for this in terms of momentum from Newton’s Second Law equation F=ma, where the force F is constant.

Remembering that velocity, force and therefore impulse are vector quantities.

For a mass m being accelerated by a constant force F, where the impulse is J , v₁ is initial velocity and v₂ is final velocity:

Ft = m(v₂– v₁)

J = Ft

J = m(v₂– v₁)

Units

Since impulse is the product of force and time, the units of impulse are (Newtons) x (seconds), or N s .

Vector problems

Vector type questions on impulse are solved by first calculating the change in momentum. This gives a vector expression for the impulse. Using Pythagoras, the magnitude of the impulse can then be found. The anglular direction is calculated from the coefficients of unit vectors i and j.

Example #1

A particle of mass 0.5 kg moves with a constant velocity of    (3i+ 5j) m.s^-1 .
After being given an impulse, the particle then moves off with a constant velocity of (2i – 3j) m.s^-1 .

Calculate:

i)   the impulse
ii)  the magnitude of the impulse( to 2 d.p.)
iii) the direction of the impulse(θ degrees to the x-axis)

i)

v₁= (3i + 5j)      v₂ = (2i – 3j)     m = 0.5 kg

using J = m(v₂– v₁)

J = 0.5(2i – 3j) – 0.5(3i + 5j)

J = i – 1.5j – 1.5i – 2.5j

J = (1 – 1.5)i + (-1.5 – 2.5)j

J = (0.5i – 4j) N.s

ii)

magnitude of impulse = √ [( 0.5)² + (-4)² ] = √ [16.23]

    = 4.03 N.s

iii) direction tan^-1 θ = (4)/(0.5) = 8

θ = 82.8749^o = 82.87^o (2 d.p.) clockwise to the x-axis

Example #2

A particle of mass 2.5 kg is moving with a constant velocity of    (2i + j) m.s^-1 .
After an impulse, the particle moves with a constant velocity of (4i + 3j) m.s^-1 .

Calculate:

i)   the impulse
ii)  the magnitude of the impulse( to 2 d.p.)
iii) the angle(θ^o) the impulse makes with the x-axis

v₁= (2i + j)     v₂ = (4i + 3j)     m = 2.5 kg

using J = m(v₂– v₁)

J = 2.5(4i +3j) – 2.5(2i + j)

J = 10i + 7.5j – 5i – 2.5j

J = (10 – 5)i + (7.5 – 2.5)j

J = (5i + 5j)

ii)

magnitude of impulse = √ [( 5)² + (5)² ] = √ [50] = 7.07 N.s

iii) direction tan^-1 θ = (5)/(5) = 1

θ = 45^o anticlockwise to the x-axis

Momentum

• The momentum of a body is equal to its mass multiplied by its velocity.

Momentum is measured in N s. Note that momentum is a vector quantity, in other words the direction is important.

Impulse

The impulse of a force (also measured in N s) is equal to the change in momentum of a body which a force causes. This is also equal to the magnitude of the force multiplied by the length of time the force is applied.

• Impulse = change in momentum = force × time

Conservation of Momentum

When there is a collision between two objects, Newton’s Third Law states that the force on one of the bodies is equal and opposite to the force on the other body.

Therefore, if no other forces act on the bodies (in the direction of collision), then the total momentum of the two bodies will be unchanged. Hence the total momentum before collision in a particular direction = total momentum after in a particular direction.

This can be used to solve problems involving colliding spheres.

Example

We have the following scenario (a ball of mass 3 kg is moving to the right with velocity 3m/s and a ball of mass 1kg is moving to the left with velocity 2m/s):

Suppose we are told that after the collision, the ball of mass 1kg moves away with velocity 2m/s, then we can use the principle of conservation of momentum to determine the velocity of the other ball after the collision.

Initial momentum = 3.3 – 2.1 = 7 [the minus sign is important: it is there because the velocity of the 1 kg body is in the opposite direction to the velocity of the 3 kg body].

Final momentum = 2 – 3x

Hence 7 = 2 – 3x (since momentum is conserved)

x = -5/3

Which means that my arrow was pointing in the wrong direction (because of the minus sign), hence the velocity of the 3 kg body after the collision is 5/3 ms^-1 to the right.

The Principle of Conservation of Momentum

The total linear momentum of a system of colliding bodies, with no external forces acting, remains constant.

for two perfectly elastic colliding bodies note:

i) By Newton’s 3rd. Law, the force on X due to Y , (F_x) is the same as the force on Y due to X , (F_y) .

F_x = F_y

ii) By Newton’s 2nd. Law, the rate of change of momentum is the same, since F = (rate of change of momentum)

iii)Because the directions of the momentum of the objects are opposite, (and therefore of different sign) the net change in momentum is zero.

Example #1

A 5 kg mass moves at a speed of 3 ms^-1 when it collides head on, with a 3 kg mass travelling at 4 ms^-1, travelling along the same line.
After the collision, the two masses move off together with a common speed.

What is the common speed of the combined masses?

Example #2

An artillery shell of mass 10 kg is fired from a field gun of mass 2000 kg.
If the speed of the shell on leaving the muzzle of the gun is 250 ms^-1 , what is the recoil speed of the gun?

Energy changes during collisions

Consider the kinetic energy change involved during a collision. Remember that no energy is actually lost, it is just converted into other forms. Energy can be transformed into heat, sound and permanent material distortion. The later causes the internal potential energy of bodies to increase.

If no kinetice energy is lost (K.E.= ½ mv² ) then the collision is said to be perfectly elastic. However if kinetic energy is lost, the collision is described as inelastic. In the special case when all the kinetic energy is lost, the collision is described as completely inelastic. This is when to two colliding bodies stick to one another on impact and have zero combined velocity.