## 2.5 impulse (ESCJK)

When a network force acts upon a body it will result in one acceleration which transforms the activity of the body. A huge net force will cause a larger acceleration than a small net force. The total change in activity of the object have the right to be the same if the big and little forces plot for various time intervals. The combination of the force and time the it action is a useful quantity i beg your pardon leads us to define impulse.

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Impulse is the product that the network force and also the time interval for which the force acts.

< extImpulse=vecF_net·Delta t>

However, indigenous Newton"s 2nd 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 change in momentum of one object. Indigenous this equation us see, the for a given change in momentum, (vecF_netDelta t) is fixed. Thus, if (vecF_net) is reduced, (Delta t) should be raised (i.e. A smaller resultant pressure must be used for much longer to bring around the same readjust in momentum). Additionally if (Delta t) is reduced (i.e. The resultant pressure is used for a much shorter period) climate the resultant force must be boosted to bring around the same adjust in momentum.

The graphs listed below show exactly how the pressure acting ~ above a body transforms with time.

The area under the graph, shaded in, to represent the advertise of the body.

A ( ext150) ( extN) resultant force acts on a ( ext300) ( extkg) trailer. Calculate exactly how long it takes this force to adjust the trailer"s velocity from ( ext2) ( extm·s\$^-1\$) come ( ext6) ( extm·s\$^-1\$) in the same direction. Assume the the forces acts to the best which is the direction of activity of the trailer.

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

The question clearly gives

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

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

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

the resultant pressure acting top top the object.

We room asked come calculate the moment taken (Delta t) to advice the trailer indigenous the ( ext2) to ( ext6) ( extm·s\$^-1\$). Native the Newton"s 2nd law,

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

Thus us have every little thing we require to discover (Delta t)!

### Choose a framework of reference

Choose appropriate as the hopeful direction.

### Do the calculation and also quote the final 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 bring away ( ext8) ( exts) for the force to change the object"s velocity from ( ext2) ( extm·s\$^-1\$) come the right to ( ext6) ( extm·s\$^-1\$) come the right.

## Worked instance 16: Impulsive cricketers

A cricket sphere weighing ( ext156) ( extg) is relocating at ( ext54) ( extkm·hr\$^-1\$) in the direction of a batsman. That is hit by the batsman ago towards the bowler at ( ext36) ( extkm·hr\$^-1\$). Calculate

the ball"s impulse, and

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

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

The question clearly gives

the ball"s mass,

the ball"s initial velocity,

the ball"s final velocity, and

the time that contact in between bat and also ball

We room asked to calculate the impulse:

< extImpulse=Delta vecp=vecF_ extnetDelta t>

Since we do not have actually the pressure exerted through the bat ~ above the sphere ( (vecF_ extnet)), we need to calculate the impulse from the readjust in momentum of the ball. Now, since

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

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

### Convert come S.I. Units

Firstly allow us adjust 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 adjust 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 framework of reference

Let us select the direction from the batsman to the bowler as the optimistic direction. Then the initial velocity of the sphere is (vecv_i=- ext15 ext m·s\$^-1\$), while the last velocity of the ball is (vecv_f=+ ext10 ext m·s\$^-1\$).

### Calculate the momentum

Now us calculate the readjust 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 indigenous the batsman to the bowler endalign*

### Determine the impulse

Finally because impulse is just the readjust in momentum of the ball,

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

### Determine the average force exerted through the bat

( extImpulse=vecF_netDelta t=Delta vecp)

We are provided (Delta t) and also we have actually calculated the advertise 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 to the bowler endalign*

## Worked example 17: Analysing a force graph

Analyse the pressure vs. Time graph listed and prize the following questions:

What is the impulse for the interval ( ext0) ( exts) to ( ext3) ( exts)? What is the impulse because that the interval ( ext3) ( exts) to ( ext6) ( exts)? What is the change in momentum for the term ( ext0) ( exts) to ( ext6) ( exts)? What is the impulse for the expression ( ext6) ( exts) come ( ext20) ( exts)? What is the impulse because that the term ( ext0) ( exts) come ( ext20) ( exts)?

### Identify what info is given and what is gift asked for

A graph of pressure versus time is provided. We space asked to identify both impulse and readjust in momentum from it.

We know that the area under the graph is the impulse and also we can relate impulse to adjust in momentum through the impulse-momentum theorem.

We need to calculate the area under the graph for the assorted intervals to determine impulse and also then occupational from there.

### Impulse for interval ( ext0) ( exts) to ( ext3) ( exts)

We should calculate the area the the shaded part under the graph. This is a triangle v a base 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) come ( ext6) ( 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 height of (- ext3) ( extN). Note that the pressure has a negative value therefore is pointing in the an unfavorable direction.

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

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

### What is the adjust in momentum for the interval ( ext0) ( exts) to ( ext6) ( exts)

From the impulse-momentum organize we recognize that the impulse is same to the adjust in momentum. Us have resolved the impulse because that the 2 sub-intervals consisting of ( ext0) ( exts) come ( ext6) ( exts). We have the right to sum lock to find the impulse because that the total interval:

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

The confident impulse in the first 3 secs is exactly opposite to the advertise in the second 3 2nd interval making the full impulse because that the first 6 secs zero:

< extimpulse_0-6 = ext0 ext N·s>

From the impulse-momentum to organize we recognize that:

### What is the impulse because that the term ( ext6) ( exts) come ( ext20) ( exts)

We have to calculate the area the the shaded part under the graph. This is split into two areas, ( ext6) ( exts) come ( ext12) ( exts) and ( ext12) ( exts) to ( ext20) ( exts), which we must 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 complete 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 advertise is ( ext2) ( extN·s) in the an unfavorable direction.

### What is the advertise of the whole period

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

## Worked example 18: automobile chase

A patrol auto is relocating on a directly horizontal road at a velocity of ( ext10) ( extm·s\$^-1\$) east. At the very same time a thief in a auto ahead of that is driving at a velocity that ( ext40) ( extm·s\$^-1\$) in the same direction.

(v_PG): velocity of the patrol vehicle relative to the ground (v_TG): velocity that the thief"s car relative to the ground

Questions 1 and also 2 native the initial version in 2011 record 1 space no longer component of the curriculum.

While travelling at ( ext40) ( extm·s\$^-1\$), the thief"s auto of massive ( ext1 000) ( extkg), collides head-on v a truck of mass ( ext5 000) ( extkg) relocating at ( ext20) ( extm·s\$^-1\$). ~ the collision, the car and also the truck relocate together. Overlook the effects of friction.

State the law of preservation of direct momentum in words.

(2 marks)

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

(6 marks)

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

Determine, by way of a calculation, even if it is the collision over could result in a deadly injury.

(5 marks)

Question 1

The full (linear) inert remains continuous (OR is conserved OR does no change) in an secluded (OR in a closed mechanism OR in the absence of external 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 ideal 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 on the car: (Taking to the ideal as positive)

eginalign* F_ extnet Delta t & = Delta p = 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 to the left as positive)

eginalign* F_ extnet Delta t & = Delta ns = 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 best 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 come the left together positive)

eginalign* F_ extnet Delta t & = Delta p = 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 on the car: (Taking to the appropriate 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 ~ above the truck: (Taking come the right 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 top top the truck: (Taking come the left as 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.