# Linear momentum

# Linear momentum

Linear momentum is a product of the body’s mass and its velocity.

The S1 unit of momentum is kgms^{-1}

When a force F is applied to a body, it changes the body’s velocity from u to v, the size of the force and the time for which it acts on a body.

From F = ma

Impulse of force is the product of force and the duration of its action or impulse is the change in momentum of the body which is acted on by the force

**Example 1**

A body of mass 3kg initially moving with a velocity of 5ms^{-1} is acted on by a horizontal force of 15N for 2s. Find the impulse and final speed.

Solution

Impulse = Ft

= 15 x 2

= 30N

Impulse = change in momentum

30 = m(v-u)

30 = 3(v-5)

v = 15ms^{-1}

**Example 2**

A tennis ball has a mass of 0.07kg. it approaches a racket with a speed of 5ms^{-1} and bounces off and returns to the way it come with a speed of 4ms^{-1}. The ball is in contact with the racket for 0.2 seconds. Calculate

(i) The impulse given to the ball

(ii) The average force exerted on the ball by the racket.

Solution

(i) Impulse = Ft = m(v-u)

= 0.07(-4-5)

= -0.63Ns

**Collisions and principles of conservation of linear momentum**

When two or more bodies collide, the total momentum of the system is conserved provided there is no external force on the system.

Consider a body of mass m_{1} moving with a velocity u_{1} to the right. Suppose the body makes a head on collision with a nother body of mass m_{2} moving with velocity u_{2} in the same direction

Let v_{1} and v_{2} be the velocities of the 2 bodies respectively after collision

Let F1 be the force exerted on m_{2} by m_{1} and F_{2} the force exerted on m_{1} by m_{2} using Newton’s 2^{nd} law.

Hence, total momentum before collision = total momentum after collision, in other words momentum is conserved.

When two bodies collide, there is a short period of contact during which each exerts a force on each other at that instant, the force which each exert on each other is equal and opposite.

Types of collision

Collisions can be categorized as inelastic collision, perfectly inelastic, elastic or perfectly elastic collisions.

Elastic or perfectly elastic collision | Inelastic collision | Perfectly inelastic collision |

Kinetic energy is conserved | Kinetic energy is not conserved | Kinetic energy is not conserved |

Linear momentum is conserved | Linear momentum is conserved | Linear momentum is conserved |

Bodies separate after collision, e.g. collision of gas molecules | Bodies separate after collision e.g. a ball bouncing from a concrete floor | Bodies stick together and move with a common velocity. E.g. a trailer colliding with a saloon car. |

Elastic collision

Momentum is conserved

m_{1}v_{1} + m_{2}v_{2} = m_{1}u_{1}+m_{2}u_{2}

m_{1}(u_{1} –v_{1}) = m_{2}(v_{2} – u_{2}) ……………………………(i)

Kinetic energy is conserved

**Example 3**

A 200g block moves to the right at a speed of 100cms^{-1} and meets a 400g block moving to the left with a speed of 80cms^{-1}. Find the final velocity of each block if the collision is elastic.

**Solution**

(v_{2} – v_{1}) = -(-0.8 – 1)

(v_{2} – v_{1}) = -1.8 …………………… (i)

Using conservation of momentum

m_{1}v_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}

(0.2 x 1) + (-0.4 x 0.8) = 0.2v_{1} + 0.4v_{2}

-0.12 = 0.2v_{1} + 0.4v_{2}

-0.6 = v_{1} + 2v_{2} …………………….(ii)

Eqn (i) and Eqn (ii)

v_{2} = -1.8 + v_{1}

-0.6 = v_{1} + 2(-1.8 +v_{1})

V_{1} = 1ms^{-1}

V_{2} = -0.8ms^{-1}

Example 4

A particle of mass m_{1}, travelling with velocity u_{1} makes a perfectly elastic collision with a stationary particle of mass m_{2}. After the collision, the first particle moves a velocity v_{1} while the second particle moves in the same direction with velocity, v_{2}. Show that

**Example 5**

A particle P of mass m_{1}, travelling with a speed u_{1} makes a head on collision with a stationary particle Q of mass m_{2}. If the collision is elastic and speed of P and Q after impact

**Example 6**

An object X of mass m moving with velocity 10ms^{-1} collides with a stationary object Y of equal mass. After collision, x, moves with speed u at an angle 30^{0} to its initial direction, while Y moves with a speed of V at an angle 90^{0} to the new direction of x.

(i) Calculate the speed u and v.

(ii) Determine whether the collision is inelastic or not.

Exercise

- A bullet of mass 300g travelling at a speed of 8ms
^{-1}hits a body of mass 450g moving in the same direction as the bullet at 1.5ms-1. The bullet and the body move together after collision. Find the loss in kinetic energy. [Ans. 3.8025J] - A particle A of mass 4kg is incident with velocity V on a stationary helium nucleus B of mass 4kg. After collision, A moves in direction BC with velocity v/2 where BC makes an angle of 600 with the initial direction of AB and the helium nucleus moves along BD. Calculate the angle made in direction AB and the velocity of the helium along BD. [θ = 30
^{0}, velocity = ]