When a net force acts on a human body it will result in an acceleration which transforms the movement of the body. A large net pressure will cause a bigger acceleration 보다 a tiny net force. The total change in activity of the object can be the same if the huge and little forces act for various time intervals. The mix of the force and time that it action is a helpful quantity which leads us to define impulse.

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Impulse is the product of the net force and the time interval because that which the force acts.

< extImpulse=vecF_net·Delta t>

However, native Newton"s second Law, we recognize that

eginalign* vecF_net& = fracDelta vecpDelta t \ herefore vecF_net·Delta t& = Delta vecp \ & = extImpulse endalign*

Therefore we can specify the impulse-momentum theorem:

< extImpulse=Delta vecp>

Impulse is same to the adjust in momentum of an object. Native this equation we see, the for a given adjust in momentum, (vecF_netDelta t) is fixed. Thus, if (vecF_net) is reduced, (Delta t) should be raised (i.e. A smaller resultant force must be applied for much longer to bring around the same change in momentum). Conversely if (Delta t) is lessened (i.e. The resultant pressure is applied for a shorter period) climate the resultant pressure must be boosted to bring around the same change in momentum.

The graphs below show just how the pressure acting on a body changes with time.  The area under the graph, shaded in, to represent the impulse of the body.

## Worked instance 15: advertise and change in momentum

A ( ext150) ( extN) result force acts on a ( ext300) ( extkg) trailer. Calculate just how long that takes this pressure to readjust the trailer"s velocity from ( ext2) ( extm·s\$^-1\$) come ( ext6) ( extm·s\$^-1\$) in the exact same direction. Assume that the pressures acts to the appropriate which is the direction of motion of the trailer.

### Identify what details is given and also what is inquiry for

The question clearly gives

the trailer"s mass together ( ext300) ( extkg),

the trailer"s initial velocity together ( ext2) ( extm·s\$^-1\$) to the right,

the trailer"s last velocity as ( ext6) ( extm·s\$^-1\$) come the right, and

the resultant force acting top top the object.

We space asked come calculate the time taken (Delta t) to accelerate the trailer indigenous the ( ext2) to ( ext6) ( extm·s\$^-1\$). Native the Newton"s second law,

eginalign* vecF_netDelta t& = Delta vecp \ & = mvecv_f-mvecv_i \ Delta t& = fracmvecF_netleft(vecv_f-vecv_i ight). endalign*

Thus us have whatever we need to uncover (Delta t)!

### Choose a frame of reference

Choose appropriate as the confident direction.

### Do the calculation and also quote the last answer

eginalign* Delta t& = fracmvecF_netleft(vecv_f-vecv_i ight) \ Delta extt& = left(frac300+150 ight)left(left(+6  ight)-left(+2  ight) ight) \ Delta extt& = left(frac300150 ight)left(4  ight) \ Delta t& = fracleft(300 ight)left(+4  ight)150 \ Delta t& = ext8 ext s endalign*

It takes ( ext8) ( exts) for the pressure to change the object"s velocity native ( ext2) ( extm·s\$^-1\$) come the right to ( ext6) ( extm·s\$^-1\$) come the right.

## Worked example 16: Impulsive cricketers

A cricket ball weighing ( ext156) ( extg) is relocating at ( ext54) ( extkm·hr\$^-1\$) towards a batsman. The is hit by the batsman earlier towards the bowler at ( ext36) ( extkm·hr\$^-1\$). Calculate

the ball"s impulse, and

the average pressure exerted by the bat if the round is in call with the bat because that ( ext0,13) ( exts).

### Identify what details is given and what is request for

The question explicitly gives

the ball"s mass,

the ball"s early velocity,

the ball"s last velocity, and

the time of contact in between bat and also ball

We are asked to calculate the impulse:

< extImpulse=Delta vecp=vecF_ extnetDelta t>

Since we do not have the pressure exerted by the bat top top the sphere ( (vecF_ extnet)), we have to calculate the impulse from the readjust in inert of the ball. Now, since

eginalign* Delta vecp& = vecp_f-vecp_i \ & = mvecv_f-mvecv_i, endalign*

we require the ball"s mass, early velocity and final velocity, i beg your pardon we room given.

### Convert come S.I. Units

Firstly permit us change units for the mass

eginalign* ext1 000  extg& = 1  extkg \ extSo, 1  extg& = frac1 ext1 000  extkg \ herefore 156 imes 1  extg& = 156 imes frac1 ext1 000  extkg \ & = ext0,156  extkg endalign*

Next we change units for the velocity

eginalign* 1  extkm· exth^-1& = frac ext1 000  extm ext3 600  exts \ herefore 54 imes 1  extkm· exth^-1& = 54 imes frac ext1 000  extm ext3 600  exts \ & = 15  extm·s\$^-1\$ endalign*

Similarly, ( ext36) ( extkm·hr\$^-1\$) = ( ext10) ( extm·s\$^-1\$).

### Choose a frame of reference

Let us select the direction native the batsman to the bowler as the optimistic direction. Climate the early stage velocity of the round is (vecv_i=- ext15 ext m·s\$^-1\$), when the last velocity that the round is (vecv_f=+ ext10 ext m·s\$^-1\$).

### Calculate the momentum

Now we calculate the adjust in momentum,

eginalign* Delta vecp& = vecp_f-vecp_i \ & = mvecv_f-mvecv_i \ & = mleft(vecv_f-vecv_i ight) \ & = left( ext0,156 ight)left(left(+10  ight)-left(-15  ight) ight) \ & = ext+3,9  \ & = ext3,9 ext kg·m·s\$^-1\$~ extin the direction native the batsman come the bowler endalign*

### Determine the impulse

Finally due to the fact that impulse is simply the adjust in momentum of the ball,

eginalign* extImpulse& = Delta vecp \ & = ext3,9 ext kg·m·s\$^-1\$~ extin the direction native the batsman to the bowler endalign*

### Determine the average pressure exerted by the bat

( extImpulse=vecF_netDelta t=Delta vecp)

We are given (Delta t) and we have actually calculated the impulse of the ball.

eginalign* vecF_netDelta t& = extImpulse \ vecF_netleft( ext0,13 ight)& = ext+3,9 \ vecF_net& = frac ext+3,9 ext0,13 \ & = +30 \ & = ext30 ext N~ extin the direction from the batsman come the bowler endalign*

## Worked example 17: Analysing a force graph

Analyse the pressure vs. Time graph noted and prize the adhering to questions:

What is the impulse because that the expression ( ext0) ( exts) come ( ext3) ( exts)? What is the impulse because that the term ( ext3) ( exts) come ( ext6) ( exts)? What is the readjust in momentum because that the expression ( ext0) ( exts) to ( ext6) ( exts)? What is the impulse for the expression ( ext6) ( exts) come ( ext20) ( exts)? What is the impulse for the interval ( ext0) ( exts) to ( ext20) ( exts)? ### Identify what details is given and also what is being asked for

A graph of force versus time is provided. We are asked to identify both advertise and change in momentum from it.

We recognize that the area under the graph is the impulse and we have the right to relate impulse to change in momentum v the impulse-momentum theorem.

We must calculate the area under the graph for the miscellaneous intervals to recognize impulse and then work from there.

### Impulse because that interval ( ext0) ( exts) to ( ext3) ( exts) We should calculate the area that the shaded portion under the graph. This is a triangle through a basic of ( ext3) ( exts) and a elevation of ( ext3) ( extN) therefore:

eginalign* extImpulse &= frac12bh \ & = frac12(3)(3) \ & = ext4,5 ext N·s endalign*

The impulse is ( ext4,5) ( extN·s) in the optimistic direction.

### Impulse for interval ( ext3) ( exts) to ( ext6) ( exts) We have to calculate the area that the shaded section under the graph. This is a triangle v a base of ( ext3) ( exts) and a elevation of (- ext3) ( extN). Keep in mind that the pressure has a negative value for this reason is pointing in the an adverse direction.

eginalign* extImpulse &= frac12bh \ & = frac12(3)(-3) \ & = - ext4,5 ext N·s endalign*

The advertise is ( ext4,5) ( extN·s) in the an unfavorable direction.

### What is the readjust in momentum because that the expression ( ext0) ( exts) come ( ext6) ( exts) From the impulse-momentum to organize we understand that that impulse is same to the change in momentum. We have worked out the impulse because that the two sub-intervals consisting of ( ext0) ( exts) come ( ext6) ( exts). We can sum them to uncover the impulse for the total interval:

eginalign* extimpulse_0-6 & = extimpulse_0-3 + extimpulse_3-6 \ &= ( ext4,5)+(- ext4,5) \ &= ext0 ext N·s endalign*

The optimistic impulse in the first 3 seconds is exactly opposite to the impulse in the 2nd 3 2nd interval make the full impulse because that the an initial 6 secs zero:

< extimpulse_0-6 = ext0 ext N·s>

From the impulse-momentum theorem we recognize that:

### What is the impulse for the term ( ext6) ( exts) come ( ext20) ( exts) We should calculate the area that the shaded section under the graph. This is divided into two areas, ( ext6) ( exts) come ( ext12) ( exts) and also ( ext12) ( exts) come ( ext20) ( exts), which we have to sum to get the complete impulse.

eginalign* extImpulse_6-12 &= (6)(-3) \ & = - ext18 ext N·s endalign*eginalign* extImpulse_12-20 &= (8)(2) \ & = + ext16 ext N·s endalign*

The full impulse is the sum of the two:

eginalign* extImpulse_6-20 &= extImpulse_6-12 + extImpulse_12-20 \ &= (-18) + (16) \ &= - ext2 ext N·s endalign*

The impulse is ( ext2) ( extN·s) in the negative direction.

### What is the impulse of the whole period The advertise is ( ext2) ( extN·s) in the an adverse direction.

## Worked example 18: auto chase

A patrol automobile is moving on a straight horizontal road at a velocity that ( ext10) ( extm·s\$^-1\$) east. In ~ the exact same time a thef in a car ahead of him is driving in ~ a velocity the ( ext40) ( extm·s\$^-1\$) in the same direction. (v_PG): velocity the the patrol auto relative to the ground (v_TG): velocity that the thief"s auto relative come the ground

Questions 1 and also 2 native the original version in 2011 paper 1 are no longer part of the curriculum.

While travel at ( ext40) ( extm·s\$^-1\$), the thief"s car of massive ( ext1 000) ( extkg), collides head-on through a truck of fixed ( ext5 000) ( extkg) moving at ( ext20) ( extm·s\$^-1\$). ~ the collision, the car and the truck relocate together. Neglect the effects of friction. State the law of preservation of linear momentum in words.

(2 marks)

Calculate the velocity the the thief"s car instantly after the collision.

(6 marks)

Research has presented that pressures greater than 85 000 N throughout collisions may cause fatal injuries. The collision described over lasts for ( ext0,5) ( exts).

Determine, by method of a calculation, whether the collision above could an outcome in a deadly injury.

(5 marks)

Question 1

The complete (linear) inert remains constant (OR is conserved OR does no change) in an secluded (OR in a closed mechanism OR in the lack of exterior forces).

(2 marks)

Question 2

Option 1:

Taking come the ideal as positive

eginalign* sum p_ extbefore & = sum p_ extafter \ ( ext1 000)( ext40)+( ext5 000)(- ext20) & = ( ext1 000+ ext5 000)v_f \ v_f & = - ext10 mcdot exts^-1\ & = ext10 mcdot exts^-1 quad extleft OR west endalign*

Option 2:

Taking to the best as positive

eginalign* Delta p_ extcar& = - Delta p_ exttruck \ m_ extcar(v_f -v_i, extcar) & = - m_ exttruck(v_f -v_i, exttruck) \ ( ext1 000)(v_f-( ext40)) & = -( ext5 000)(v_f -(- ext20)) \ ext6 000v_f & = - ext60 000 \ herefore v_f & = - ext10 mcdot exts^-1 \ herefore v_f & = ext10 mcdot exts^-1 quad extleft OR west endalign*

(6 marks)

Question 3

Option 1:

Force ~ above the car: (Taking to the appropriate as positive)

eginalign* F_ extnet Delta t & = Delta ns = mv_f - mv_i \ F_ extnet ( ext0,5) & = ( ext1 000)(- ext10- ext40) \ herefore F_ extnet & = -10^5 ext N \ & extOR \ herefore F_ extnet & = 10^5 ext N quad ( ext100 000 N)\ herefore F_ extnet & > ext85 000 N endalign*

Yes, the collision is fatal.

OR

Force top top the car: (Taking come the left as positive)

eginalign* F_ extnet Delta t & = Delta p = mv_f - mv_i \ F_ extnet ( ext0,5) & = ( ext1 000)( ext10-(-40)) \ herefore F_ extnet & = 10^5 ext N quad ( ext100 000 N)\ herefore F_ extnet & > ext85 000 N endalign*

Yes, the collision is fatal.

Option 2:

Force ~ above the truck: (Taking to the right as positive)

eginalign* F_ extnet Delta t & = Delta p = mv_f - mv_i \ F_ extnet ( ext0,5) & = ( ext5 000)(-10-(-20)) \ herefore F_ extnet & = 10^5 ext N quad ( ext100 000 N)\ herefore F_ extnet & > ext85 000 N endalign*

Yes, the collision is fatal.

OR

Force top top the truck: (Taking to the left as positive)

eginalign* F_ extnet Delta t & = Delta ns = mv_f - mv_i \ F_ extnet ( ext0,5) & = ( ext1 000)( ext10- ext20) \ herefore F_ extnet & = -10^5 ext N \ & extOR \ herefore F_ extnet & = 10^5 ext N quad ( ext100 000 N)\ herefore F_ extnet & > ext85 000 N endalign*

Yes, the collision is fatal.

Option 3:

Force ~ above the car: (Taking come the ideal as positive)

eginalign* v_f & = v_i + a Delta t \ -10 & = ext40 + a ( ext0,5)\ herefore a & = - ext100 mcdot exts^-2\ F_ extnet & = ma \ & = ( ext1 000)(-100) \ F_ extnet & = -10^5 ext N quad (- ext100 000 N)\ F_ extnet & = 10^5 ext N quad ( ext100 000 N)\ herefore F_ extnet & > ext85 000 N endalign*

Yes, the collision is fatal.

OR

Force top top the car: (Taking come the left as positive)

eginalign* v_f & = v_i + a Delta t \ ext10 & = -40 + a ( ext0,5)\ herefore a & = ext100 mcdot exts^-2\ F_ extnet & = ma \ & = ( ext1 000)( ext100) \ F_ extnet & = 10^5 ext N quad ( ext100 000 N)\ herefore F_ extnet & > ext85 000 N endalign*

Yes, the collision is fatal.

Option 4:

Force top top the truck: (Taking to the best as positive)

eginalign* v_f & = v_i + a Delta t \ -10 & = -20 + a ( ext0,5)\ herefore a & = ext20 mcdot exts^-2\ F_ extnet & = ma \ & = ( ext5 000)( ext20) \ F_ extnet & = 10^5 ext N quad ( ext100 000 N)\ herefore F_ extnet & > ext85 000 N endalign*

Yes, the collision is fatal.

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OR

Force ~ above the truck: (Taking come the left together positive)

eginalign* v_f & = v_i + a Delta t \ ext10 & = ext20 + a ( ext0,5)\ herefore a & = - ext20 mcdot exts^-2\ F_ extnet & = ma \ & = ( ext5 000)(-20) \ F_ extnet & = -10^5 ext N quad (- ext100 000 N)\ F_ extnet & = 10^5 ext N quad ( ext100 000 N)\ herefore F_ extnet & > ext85 000 N endalign*

Yes, the collision is fatal.