11.10 Impulse

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Impulse

Impulse is the product of the net force and the time interval for which the force acts. Impulse is defined as:

Impulse = F \cdot Δ t

However, from Newton's Second Law, we know that

\begin{matrix} F & = & \frac{Δ p}{Δ t} \\ ∴ F \cdot Δ t & = & Δ p \\ = & Impulse \end{matrix}

Therefore,

Impulse = Δ p

Impulse is equal to the change in momentum of an object. From this equation we see, that for a given change in momentum, $F_{n e t} Δ t$ is fixed. Thus, if $F_{n e t}$ is reduced, $Δ t$ must be increased (i.e. a smaller resultant force must be applied for longer to bring about the same change in momentum). Alternatively if $Δ t$ is reduced (i.e. the resultant force is applied for a shorter period) then the resultant force must be increased to bring about the same change in momentum.

A 150 N resultant force acts on a 300 kg trailer. Calculate how long it takes this force to change the trailer's velocity from 2 m $\cdot$ s $^{- 1}$ to 6 m $\cdot$ s $^{- 1}$ in the same direction. Assume that the forces acts to the right.

Identify what information is given and what is asked for
The question explicitly gives
- the trailer's mass as 300 kg,
- the trailer's initial velocity as 2 m $\cdot$ s $^{- 1}$ to the right,
- the trailer's final velocity as 6 m $\cdot$ s $^{- 1}$ to the right, and
- the resultant force acting on the object
all in the correct units!

We are asked to calculate the time taken $Δ t$ to accelerate the trailer from the 2 to 6 m $\cdot$ s $^{- 1}$ . From the Law of Momentum,

$\begin{matrix} F_{n e t} Δ t & = & Δ p \\ = & m v_{f} - m v_{i} \\ = & m (v_{f} - v_{i}) . \end{matrix}$

Thus we have everything we need to find $Δ t$ !
Choose a frame of reference
Choose right as the positive direction.
Do the calculation and quote the final answer
$\begin{matrix} F_{n e t} Δ t & = & m (v_{f} - v_{i}) \\ (+ 150 N) Δ t & = & (300 kg) ((+ 6 m \cdot s^{- 1}) - (+ 2 m \cdot s^{- 1})) \\ (+ 150 N) Δ t & = & (300 kg) (+ 4 m \cdot s^{- 1}) \\ Δ t & = & \frac{(300 kg) (+ 4 m \cdot s^{- 1})}{+ 150 N} \\ Δ t & = & 8 s \end{matrix}$

It takes 8 s for the force to change the object's velocity from 2 m $\cdot$ s $^{- 1}$ to the right to 6 m $\cdot$ s $^{- 1}$ to the right.

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A cricket ball weighing 156 g is moving at 54 km $\cdot$ hr $^{- 1}$ towards a batsman. It is hit by the batsman back towards the bowler at 36 km $\cdot$ hr $^{- 1}$ . Calculate

the ball's impulse, and
the average force exerted by the bat if the ball is in contact with the bat for 0,13 s.

Identify what information is given and what is asked for
The question explicitly gives
- the ball's mass,
- the ball's initial velocity,
- the ball's final velocity, and
- the time of contact between bat and ball
We are asked to calculate the impulse

$Impulse = Δ p = F_{n e t} Δ t$

Since we do not have the force exerted by the bat on the ball (F $_{n e t}$ ), we have to calculate the impulse from the change in momentum of the ball. Now, since

$\begin{matrix} Δ p & = & p_{f} - p_{i} \\ = & m v_{f} - m v_{i}, \end{matrix}$

we need the ball's mass, initial velocity and final velocity, which we are given.
Convert to S.I. units
Firstly let us change units for the mass

$\begin{matrix} 1000 g & = & 1 kg \\ So, 1 g & = & \frac{1}{1000} kg \\ ∴ 156 \times 1 g & = & 156 \times \frac{1}{1000} kg \\ = & 0, 156 kg \end{matrix}$

Next we change units for the velocity

$\begin{matrix} 1 km \cdot h^{- 1} & = & \frac{1000 m}{3 600 s} \\ ∴ 54 \times 1 km \cdot h^{- 1} & = & 54 \times \frac{1 000 m}{3 600 s} \\ = & 15 m \cdot s^{- 1} \end{matrix}$

Similarly, 36 km $\cdot$ hr $^{- 1}$ = 10 m $\cdot$ s $^{- 1}$ .
Choose a frame of reference
Let us choose the direction from the batsman to the bowler as the positive direction. Then the initial velocity of the ball is $v_{i}$ = -15 m $\cdot$ s $^{- 1}$ , while the final velocity of the ball is $v_{f}$ = 10 m $\cdot$ s $^{- 1}$ .
Calculate the momentum
Now we calculate the change in momentum,

$\begin{matrix} p & = & p_{f} - p_{i} \\ = & m v_{f} - m v_{i} \\ = & m (v_{f} - v_{i}) \\ = & (0, 156 kg) ((+ 10 m \cdot s^{- 1}) - (- 15 m \cdot s^{- 1})) \\ = & + 3, 9 kg \cdot m \cdot s^{- 1} \\ = & 3, 9 kg \cdot m \cdot s^{- 1} in the direction from batsman to bowler \end{matrix}$
Determine the impulse
Finally since impulse is just the change in momentum of the ball,

$\begin{matrix} Impulse & = & Δ p \\ = & 3, 9 kg \cdot m \cdot s^{- 1} in the direction from the batsman to the bowler \end{matrix}$
Determine the average force exerted by the bat
$Impulse = F_{n e t} Δ t = Δ p$

We are given $Δ t$ and we have calculated the impulse of the ball.

$\begin{matrix} F_{n e t} Δ t & = & Impulse \\ F_{n e t} (0, 13 s) & = & + 3, 9 N \cdot s \\ F_{n e t} & = & \frac{+ 3, 9 N \cdot s}{0, 13 s} \\ = & + 30 N \\ = & 30 N in the direction from batsman to bowler \end{matrix}$

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Exercise

Which one of the following is NOT a unit of impulse?
1. A $N \cdot s$
2. B $k g \cdot m \cdot s^{- 1}$
3. C $J \cdot m \cdot s^{- 1}$
4. D $J \cdot m^{- 1} \cdot s$
A toy car of mass 1 kg moves eastwards with a speed of 2 m · s - 1 . It collides head-on with a toy train. The train has a mass of 2 kg and is moving at a speed of 1,5 m · s - 1 westwards. The car rebounds (bounces back) at 3,4 m · s - 1 and the train rebounds at 1,2 m · s - 1 .
1. Calculate the change in momentum for each toy.
2. Determine the impulse for each toy.
3. Determine the duration of the collision if the magnitude of the force exerted by each toy is 8 N.
A bullet of mass 20 g strikes a target at 300 m · s - 1 and exits at 200 m · s - 1 . The tip of the bullet takes 0,0001s to pass through the target. Determine:
1. the change of momentum of the bullet.
2. the impulse of the bullet.
3. the magnitude of the force experienced by the bullet.
A bullet of mass 20 g strikes a target at 300 m · s - 1 . Determine under which circumstances the bullet experiences the greatest change in momentum, and hence impulse:
1. When the bullet exits the target at 200 m $\cdot$ s $^{- 1}$ .
2. When the bullet stops in the target.
3. When the bullet rebounds at 200 m $\cdot$ s $^{- 1}$ .
A ball with a mass of 200 g strikes a wall at right angles at a velocity of 12 m · s - 1 and rebounds at a velocity of 9 m · s - 1 .
1. Calculate the change in the momentum of the ball.
2. What is the impulse of the wall on the ball?
3. Calculate the magnitude of the force exerted by the wall on the ball if the collision takes 0,02s.
If the ball in the previous problem is replaced with a piece of clay of 200 g which is thrown against the wall with the same velocity, but then sticks to the wall, calculate:
1. The impulse of the clay on the wall.
2. The force exerted by the clay on the wall if it is in contact with the wall for 0,5 s before it comes to rest.

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Source: OpenStax, Siyavula textbooks: grade 11 physical science. OpenStax CNX. Jul 29, 2011 Download for free at http://cnx.org/content/col11241/1.2

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