Momentum and Collisions

Momentum is a physical quantity used as a measure of the amount of motion an object has. It is directly proportional to the mass and the velocity of the object. It is defined to be the product of the mass and velocity of the object. Since velocity is a vector quantity, momentum is also a vector quantity.

p = mv

p is the momentum of an object of mass m moving with a velocity v. This is a vector equation. To obtain a relationship between the components along a certain direction, components of both sides along that direction should be taken.

p = mv

p and v are the components of the momentum and the velocity vectors respectively. The unit of measurement for momentum is kg m ⁄ s.

Example: Calculate the momentum of a 6 kg object moving with a speed of 4 m ⁄ s.

Solution: m = 6 kg ; v = 4 m ⁄ s ; p = ?

p = mv = 6 * 4 kg m ⁄ s = 24 kg m ⁄ s

Impulse of an object is defined to be the product of the force acting on the object and the time interval during which the force is applied. Impulse is a vector quantity. The unit of measurement for impulse is N s.

I = FΔt

I is impulse of an object acted upon by a force F for a time interval of Δt. This is a vector equation. A component equation may be obtained by equating the components of both sides.

I = FΔt

Example: An object was acted upon by a force of 50 N for 0.2 seconds. Calculate the impulse acting on the object.

Solution: F = 50 N ; Δt = 0.2 s ; I = ?

I = FΔt = 50 * 0.2 N s = 10 N s

Impulse can be obtained from a graph of force versus time as the area enclosed between the force versus time curve and the time axis.

Principle of Conservation of Momentum

According to Newton's second law, force is equal to the product of mass and acceleration and acceleration is equal to the rate of change of velocity with time.

F = mΔv ⁄ Δt = ( mv_f - mv_i ) ⁄ Δt

But mv_i = p_i and mv_f = p_f.

F = ( p_f - p_i ) ⁄ Δt = Δp ⁄ Δt

The relationship between force and momentum is that force acting on an object is equal to the rate of change of its momentum with time. This relationship also implies that the change in the momentum of an object is equal to the product of force and interval of time which is in turn equal to the impulse acting on the object.

I = Δp

Example: A ball of mass 0.1 kg moving to the right hits a wall with a speed of 10 m ⁄ and bounces back with a speed of 6 m ⁄ s.

Calculate the change in its momentum.

Solution: Remember a component to the right is taken to be positive and a component to the left is taken to be negative.

m = 0.1 kg ; v_i = 10 m ⁄ s ; v_f = -6 m ⁄ s ; Δp = ?
Δp = p_f - p_i = mv_f - mv_i = (0.1 * -6 - 0.1 * 10) kg m ⁄ s ; = -1.6 kg m ⁄ s

Calculate the impulse impacted on the object during the collision.

Solution: I = ?
I = Δp = -1.6 N s

If the ball was in contact with the wall for 0.2 seconds, Calculate the average force exerted by the wall on the ball.

Solution: Δt = 0.2 s ; F = ?
F = Δp ⁄ Δt = -1.6 ⁄ 0.2 N s = -8 N s
The negative means the direction of the force is to the left.

If the net force acting on an object is zero, then Δp = FΔt = 0. That is the momentum of the object does not change. In other words the momentum of the object is conserved. The principle of conservation of momentum states that if the net force acting on an object is zero, then the momentum of the object is conserved.

F = 0

then

p_f = p_i

Conservation of Momentum for Collisions

When two objects collide, each object exerts force on the other. These two forces are action reaction forces. They are opposites of each other. If the two objects are treated as a single system, these two forces cancel each other and the net force acting on the system during the collision is zero. And from the principle of conservation of momentum, it follows that momentum of colliding objects is conserved if they are treated as a single system.

One Dimensional Collision

One dimensional collision is collision in a straight line. Suppose an object of mass m₁ moving with a speed v_1i collides with an object of mass m₂ moving with a speed of v_2i. And suppose after collision, the former moves with a speed of v_1f and the later moves with a speed of v_2f. Since momentum of colliding objects is conserved, the momentum of these colliding objects before collision is equal to their momentum after collision.

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

Example: A 10 kg object moving to the right with a speed of 5 m ⁄ s collides with a 2 kg object moving to the left with a speed of 3 m ⁄ s. After collision, the 2 kg object moves to the right with a speed of 10 m ⁄ s. Calculate the speed of the 10 kg object after collision.

Solution: m₁ = 10 kg ; v_1i = 5 m ⁄ s ; m₂ = 2 kg ; v_2i = -3 m ⁄ s ; v_2f = 10 m ⁄ s ; v_1f = ?

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

m₁v_1f = m₁v_1i + m₂v_2i - m₂v_2f = ( 10 * 5 + 2 * -3 - 2 * 10 ) kg m ⁄ s = 24 kg m ⁄ s

v_1f = 24 ⁄ 10 m ⁄ s = 2.4 m ⁄ s

Completely Inelastic Collision

A Completely inelastic collision is a collision where the two objects stick together after collision. Both objects will have the same speed after collision: v_1f = v_2f = V.

m₁v_1i + m₂v_2i = ( m₁ + m₂ )V

V is their common speed after collision. Kinetic energy is not conserved during inelastic collision. Some of the initial kinetic energy is lost as heat energy during the collision.

Example: A 7 kg object moving to the right with a speed of 8 m ⁄ s collides with a 3 kg object moving to the left with a speed of 5 m ⁄ s. If the collision is completely inelastic

Calculate their speed after the collision.

Solution: m₁ = 7 kg ; v_1i = 8 m ⁄ s ; m₂ = 3 kg ; v_2i = -5 m ⁄ s ; V = ?
m₁v_1i + m₂v_2i = ( m₁ + m₂ )V

V = ( m₁v_1i + m₂v_2i ) ⁄ (m₁ + m₂ ) = ( 7 * 8 + 3 * -5) ⁄ ( 7 + 3 ) m ⁄ s = 4.1 m ⁄ s

Calculate the kinetic energy lost during the collision.

Solution: ΔKE = KE_f - KE_i = ?
ΔKE = ( m₁ + m₂ )V² ⁄ 2 ) - (m₁v_1i² ⁄ 2 + m₂v_2i² ⁄ 2 )

= { ( 7 + 3 )* 4.1² ⁄ 2 ) - ( 7 * 8² ⁄ / 2 - 3 * -5² ⁄ 2 ) } J

= J