Mechanics A level question needing help with?

[Fedex16]

Can anyone provide a solution to this A Level dynamics question please?

" A gun of mass 600kg is free to move along a horizontal track and is connected by a light inelastic rope to an open truck containing sand whose total mass is 1490kg. The truck is free to move along the same track as the gun. A shell of mass 10kg is fired from the gun towards the truck and when it leaves the barrel has a horizontal velocity of 915ms-1 relative to the gun and parallel to the track. The shell lodges in the sand where it comes to relative rest before the rope tightens. Find:

1.)The speeds of the gun and the shell just after the shell leaves the barrel,
2.)The speed of the truck before the rope tightens when the shell is at relative rest inside the truck,
3.)The speed of the gun and truck just after the rope tightens,
4.)The loss in kinetic energy due to the rope tightening,
5.)The magnitude of the impulsive tension in the rope."

Many thanks
Fedex16

 

[wjm11]

" A gun of mass 600kg is free to move along a horizontal track and is connected by a light inelastic rope to an open truck containing sand whose total mass is 1490kg. The truck is free to move along the same track as the gun. A shell of mass 10kg is fired from the gun towards the truck and when it leaves the barrel has a horizontal velocity of 915ms-1 relative to the gun and parallel to the track. The shell lodges in the sand where it comes to relative rest before the rope tightens. Find:

1.)The speeds of the gun and the shell just after the shell leaves the barrel,
2.)The speed of the truck before the rope tightens when the shell is at relative rest inside the truck,
3.)The speed of the gun and truck just after the rope tightens,
4.)The loss in kinetic energy due to the rope tightening,
5.)The magnitude of the impulsive tension in the rope."

1.)The speeds of the gun and the shell just after the shell leaves the barrel:

Conservation of Momentum. Before the gun is fired, the momentum is zero. The firing of the gun is done by internal forces; no external forces have been applied. Therefore, there is no change in momentum. You know the masses of both objects and the difference in their velocities. The mv of one must be equal and opposite to the mv of the other.

2.)The speed of the truck before the rope tightens when the shell is at relative rest inside the truck:

Conservation of Momentum. From part 1, you know the velocity of the shell, so you can find the momentum of the shell-truck system. Once again, there are no external forces applied, so the momentum before impact must equal the momentum after impact.

3.)The speed of the gun and truck just after the rope tightens:

Conservation of Momentum. We are back where we started before the shell was fired – zero momentum – so both gun and truck must be at rest.

4.)The loss in kinetic energy due to the rope tightening:

You have already found the velocities of the gun and truck in parts 1 and 2. Use this to find kinetic energies, (1/2)mv^2. After the rope tightens, the vehicles return to rest and their KE is zero.

5.)The magnitude of the impulsive tension in the rope:

This cannot be determined from the information provided. The impulse (change in momentum) can be found, but it is the product of the force (tension in the rope) and the duration (time) that the force is applied. Since the duration is not given, the force cannot be calculated.

 

[Fedex16]

Gun and truck- have attempted qu.

Thanks for your help. I have looked at some notes and have a possible solution , what do you think?

Fedex 16
Gun and truck

1. Speed of gun and shell

Let Vg = speed of gun, Vs = speed of shell, Vt = speed of truck, sVg = speed of shell relative to gun
From relative motion: sVg = Vs – Vg → Vs = sVg – Vg (1)
Using Conservation of linear momentum
Initial speeds are zero → 0 = MgVg + MsVs (2)
From (1) → MgVg + Ms (sVg-Vg)
→ -600*Vg + 1490 (915 – Vg)
→ Vg = -91500/590 = -155 ms-1
From (1) Vs = 915- (-155) = 1070 ms-1

1. Velocity truck before rope tightens-ignore effect of gun as rope is still slack

Vt0 = Speed truck initially, Vt1 = Speed truck after shell lodges in sand
Using COM (Mt Vt0) + (VsMs) = (Ms + Mt) Vt1
→ 0 + (10 * 1070) = (10 + 1490) Vt1
→ Vt1 = 10 * 1070/ (10+1490) = 7.133 ms-1

1. Speed gun and truck just after rope tightens

Using COM, M1V1 + M2V2 = M3V3 + M4V4, but here V3 =V4 assuming they both go off in the same direction
→ (600 * -155) + (1500 * 7.133) = (1500 + 600) Vgt where Vgt = Speed of gun and truck after rope tightens
→ Vgt = 10700 -93000/ 2100 = -39.20 ms-1 in direction of gun

1. Loss of KE due to rope tightening:

KEg = 0.5 MgVg^2 = 0.5 * 600 * 155^2 = 7,207,500
KEt = 0.5 * 1500 * 7.133^2 = 38,160
Total KE before = 7245660 J
KE after = 0.5 (600 + 1500) * 39.2^2 = 1,613,472 J
→ Loss KE = (7245660 – 1613472) = 5,632,188 J

1. Impulsive tension in rope

The jerk of the rope creates an impulsive tension on the truck rope which is equal and opposite to that of the gun rope.
At truck end, Change in momentum = Impulsive tension in absence of any other info ie duration:
MV before = 1500 * 7.133 = 10700
MV after = 1500* -39.20 = -58800
→ IT = Mu –Mv = 10700 – (-58800) = 82300 N

Wjm11
I have looked again at the question which was taken straight off an exam paper and there are no errors in the wording. Obviously I am concerned about the Impulsive tension- is it really a red herring. I mean are they expecting us to say that it is incalculable. Your thoughts on above would be appreciated.

I'm still very much learning about this site. Tried to upload a word doc but was told its 6kb too big so have copied to clipboard and pasted above. Hope it is discernible.


Kind regards

Fedex16

 

[Deleted member 4993]

1. Impulsive tension in rope

The jerk of the rope creates an impulsive tension on the truck rope which is equal and opposite to that of the gun rope.
At truck end, Change in momentum = Impulsive tension in absence of any other info ie duration:
MV before = 1500 * 7.133 = 10700
MV after = 1500* -39.20 = -58800

The change in momentum is NOT equal to force

→ IT = Mu –Mv = 10700 – (-58800) = 82300 N * sec


Kind regards

Fedex16

.

 

[wjm11]

Let Vg = speed of gun, Vs = speed of shell, Vt = speed of truck, sVg = speed of shell relative to gun
From relative motion: sVg = Vs – Vg → Vs = sVg – Vg (1)
Using Conservation of linear momentum
Initial speeds are zero → 0 = MgVg + MsVs (2)
From (1) → MgVg + Ms (sVg-Vg)
→ -600*Vg + 1490 (915 – Vg)
→ Vg = -91500/590 = -155 ms-1
From (1) Vs = 915- (-155) = 1070 ms-1

Mass of Shell, Ms = 1490 kg ???

I think you may want to run through these numbers again.

 

[Fedex16]

Re Large Mass!

Hi, thanks for looking at this. It would appear to be a typo on this occasion. The 1490kg mass of truck was inserted by mistake instead of the 10kg mass of the shell. As I was copying from my workings the answer is as if the amount was 10kg.

ie it shows 1490 (915 + Vg) = 91500 +10Vg but should have shown 10 (915 + Vg) = 91500 + 10Vg.

Hope this makes sense

Best wishes

 

[wjm11]

1. Speed of gun and shell

Let Vg = speed of gun, Vs = speed of shell, Vt = speed of truck, sVg = speed of shell relative to gun
From relative motion: sVg = Vs – Vg → Vs = sVg – Vg (1)
Using Conservation of linear momentum
Initial speeds are zero → 0 = MgVg + MsVs (2)
From (1) → MgVg + Ms (sVg-Vg)
→ -600*Vg + 1490 (915 – Vg)
→ Vg = -91500/590 = -155 ms-1
From (1) Vs = 915- (-155) = 1070 ms-1

“sVg = Vs – Vg → Vs = sVg – Vg (1)” NO. Must ADD Vg to both sides to isolate Vs.

Vs = sVg + Vg

(590 kg)(Vg) + (10 kg)( sVg + Vg) = 0
(590 kg)(Vg) + (10 kg)(Vg) = -(10 kg)(sVg)
600Vg = -9150sVg
Vg = -15.25 m/s
Vs = -15.25 + 915 = 899.75 m/s

Check momentum:
Ps = (899.75)(10) = 8997.5 kg-m/s
Pg = (-15.25)(590) = -8997.5 kg-m/s

Always check your results before proceeding to the next step.

 

[Fedex16]

1.) Using COM as before 0= 10(sVg +Vg) - 600Vg, → 9150 = -600Vg, Vg = -15.25ms^-1.
Hence Vs = 915 +(-15.25) = 899.75ms-1.

2.) Vt. 0 + (10*899.75) = (10+1490)V_t. → V_t= 8997.5/1500 = 6ms-1.
3.) V_g and V_t just after rope tightens, ( 600* -15.25) + (1500*6) = 2100V_gt → V_gt = -0714ms^-1.
4.) Loss KE: Before 0.5*600(-15.25)²= 69768.75 (gun), 0.5*1500*6² =27000. Total before = 96768.75j.
After 0.5 (600+1500)*(-.0714)² = 5.35j. → Loss KE = (96768.75- 5.35) = 96763.40j.
5.) Impulsive tension in rope equates to change in MV at either end ie gun or truck.

Truck Before 1500*6= 9000, after 1500*(-0714)=-107.10Kgms^-1. Change = 9000+107.10=9107.10kgms^-1.
Gun before 600*15.25=9150, after 600*(0714)= 42.84kgms^-1 Change= 9150-(42.84)=9107.16kgms^-1.

I'm not sure about this last bit because it asks for impulsive tension. Impulse implies some time element is involved(not given) but Qu it is asking for the magnitude of impulse tension which seem to imply just force. Your advice would be appreciated.

Many thanks

Fedex 16

 

[wjm11]

1.) Using COM as before 0= 10(sVg +Vg) - 600Vg, → 9150 = -600Vg, Vg = -15.25ms^-1.
Hence Vs = 915 +(-15.25) = 899.75ms-1.

2.) Vt. 0 + (10*899.75) = (10+1490)V_t. → V_t= 8997.5/1500 = 6ms-1.
3.) V_g and V_t just after rope tightens, ( 600* -15.25) + (1500*6) = 2100V_gt → V_gt = -0714ms^-1.
4.) Loss KE: Before 0.5*600(-15.25)²= 69768.75 (gun), 0.5*1500*6² =27000. Total before = 96768.75j.
After 0.5 (600+1500)*(-.0714)² = 5.35j. → Loss KE = (96768.75- 5.35) = 96763.40j.
5.) Impulsive tension in rope equates to change in MV at either end ie gun or truck.

Truck Before 1500*6= 9000, after 1500*(-0714)=-107.10Kgms^-1. Change = 9000+107.10=9107.10kgms^-1.
Gun before 600*15.25=9150, after 600*(0714)= 42.84kgms^-1 Change= 9150-(42.84)=9107.16kgms^-1.

I'm not sure about this last bit because it asks for impulsive tension. Impulse implies some time element is involved(not given) but Qu it is asking for the magnitude of impulse tension which seem to imply just force. Your advice would be appreciated.

Numbers 1) and 2) are correct.

Number 3) is not correct. Focus on basic principles: Conservation of momentum MUST be maintained (at zero). Both gun and truck must come to rest after the rope tightens. Your error is in your gun mass; it should be 590 kg, not 600 kg.

Adjust the gun mass value in 4) as well as the final velocity (0).

For 5) I cannot offer you anything further than my prior comments. I would simply state the “impulse”, the change in momentum, as the answer.