__What is a impulse?__

Let us suppose that we apply a force F to an object for Δt seconds. What does the quantity F **·** Δt represent?

According to Newton's second law, the net force F on any object is related to its acceleration a and its mass m as follows:

F = m a , a is the acceleration

a = Δv / Δt = (vf - vi) / Δt , acceleration is the change of velocity over time. vf is the final velocity and vi is the initial velocity. Δt is the interval of time over which the velocity changes from vi to vf.

F = m (Δv / Δt)

From the above, we can write

F Δt = m Δv = m (vf - vi) = m vf - m vi = pf - pi

F Δt represents the change in momentum and is called the impulse of the force F for the time interval Δt.

A force F applied to an object for a time Δt gives rise to a change in momentum of the object.

Meaning of the impulse defined above

You need to stop a 0.4 Kg ball that has a speed of 20 m/s. What force is needed to stop this ball in

a) 1 s

b) 0.1 s

c) 0.01 s

Solutions

F Δt = m Δv = m (vf - vi) , we need to stop the ball, hence vf = 0, vi = 20 m/s. (assume moving in the positive direction, vi = + 20 m/s)

F Δt = 0.4(0 - 20) = - 8 Kg m/s , the minus sign indicates that F has the direction opposite the motion of the ball.

F = - 8 / Δt

a) F = - 8 / 1 = - 8 Newtons

b) F = - 8 / 0.1 = - 80 Newtons

c) F = - 8 / 0.01 = - 800 Newtons

Stopping the same ball at a speed of 20 m/s but for different time intervals requires different forces which increases as the time interval decreases. So if you allow the time interval to be larger, you need a small force to change the momentum of an object.

__Example 1__

A 2000-kilogram car traveling with a velocity of 25 m/s is stopped in 10 seconds using a breaking force F. What is the magnitude of force F?

__Solution to Example 1__

initial velocity: vi = 25 m/s (assumed to be in the positive direction)

final velocity: vf = 0 (car stops)

pi = initial momentum = m vi

pf = final momentum = m vf

|F| Δt = |pf - pi|

pf = 2000 × 0 = 0

pi = 2000 × 25 = 50000 Kg m/s

|F| = |0 - 50000| / Δt = 50000 / 10 = 5000 Newtons

__Example 2__

A football player kicks a 350-grams ball (at rest) and gives it an initial speed of 28 m/s. Find the magnitude of average force exerted by the player if the impact time is 12 ms.

__Solution to Example 2__

|F| Δt = |pf - pi|

pf = m vf = 0.35 × 28 = 9.8 (vf = 28 m/s)

pi = m vi = 0.35 × 0 = 0 (at rest vi = 0)

|F| = |9.8 - 0| / Δt = 9.8 / (12 × 10^{-3}) ≈ 817 Newtons

__Example 3__

A 1 Kg ball hits a wall with a velocity of 10 m/s to the left. The ball hits the wall and bounces off with a velocity of 8 m/s to the right. What is the average force exerted by the wall on the ball if the ball is in contact with wall for 0.1 s?

__Solution to Example 3__

Let the direction from left to right be the positive direction. The ball is moving either to the left or to the right and therefore all vector quantities such as velocities and momenta have one component only which we use in our calculations below.

vi = - 10 m/s in negative direction, vf = 8 m/s in positive direction

F Δt = pf - pi

pf = m vf = 1 × 8 = 8

pi = m vi = 1 × (- 10) = - 10

F = ( 8 - (-10)) / Δt = 18 / 0.1 = 180 Newtons , direction to the right.

__Example 4__

The drivers of two trucks traveling at the same speed, with truck A having a mass double that of truck B, apply equal forces on the brakes of the trucks to stop them. Which truck takes a longer distance to stop?

__Solution to Example 4__

Let m be the mass of truck B and 2m the mass of truk A (double)

Let initial speed be vi. The final seed is equal to 0 (trucks stop).

Truck A: F Δt = pf - pi = 0 - 2 m vi

Δt = - 2m vi / F

Truck B: pf = F Δt = pf - pi = 0 - m vi

Δt = - m vi / F

Note that F is negative because it is directed in the opposite direction of vi (considered as positive)

Comparing the period of time Δt needed to stop the trucks; truck A takes more time to stop because its momentum is larger (mass is double that of truck B) than the momentum of truck B. In fact the time taken by truck A is double the time taken by truck B to stop and therefore the stopping distance of truck A is also double the stopping distance of truck B.

More References

1 - Higher Level Physics - IB Diploma - Chris Hamper - Pearson

2 - Physics - Raymond A. Serway and Jerry S. Faughn

Holt, Reinehart and Winston - Harcourt Education Company