Mechanics: Motion of Two Connected Vehicles

[AN630078]

Homework Statement

Hello, I have been practising dynamics problems in Mechanics and found the question below while revising. Typically I am very uncertain of these questions and often end up confusing myself over minor details, would anyone be able to comment on my workings to suggest how I could improve especially when faced with similar problems or simply to find any mistakes I have overlooked?


A car of mass 1000 kg is pulling a caravan of mass 1500 kg. The car and caravan are connected by a light towbar. The total resistive forces on the car and caravan are 150 N and 200 N, respectively (throughout the motion).
The car is travelling at 20𝑚𝑠^−1 when it brakes, so as to decelerate at 0.5𝑚𝑠^−2.
Find
(i) the braking force
(ii) the tension or compression in the towbar
(iii) the distance travelled by the car and caravan before coming to rest
(iv) the time taken to come to rest

Relevant Equations

F=ma

i. Using Newton's 2nd Law, F = m a
consider the motion of the entire system, so the car, caravan and towbar an be thought of as a single object.
The tension can ignored as it is an internal force.
Braking fore + resistive forces = mass * acceleration
Braking force + 200N +150 N=(1000+1500)*(0.5)
Braking force +350N=1250 N
Braking force = 900N

Would this be correct or would it actually be -900N as when the car brakes it decelerates, i.e. a=-0.5ms^-2?

ii. Consider the forces acting on the caravan:
T=tension in the towbar
T-200=(1500)(-0.5)
T-200=-750
T=-550N

Which I think would be a compression but I am not sure why.

iii. Using suvat;
s=?
u=20
v=0
a=-0.5
t=?
Therefore, use v^2=u^2+2as
0^2=20^2+2*(-0.5)s
0=400+(-1)s
s=400m

iv. v=u+at
0=20+(-0.5)t
-20/-0.5=40 seconds

Use s=ut+1/2at^2 to check;
s=20*40+1/2*-0.5*40^2
s=800+(-400)
s=400m

Would this be correct? I have tried to comprehensively answer the question I just feel a little uncertain, particularly when it comes to dynamics problems, I worry that I may have missed something.

 

[Lnewqban]

Your calculations for i) and for ii) seem to be corrrect.

As long as you know what is happening and not doing a vectorial addition, you can keep the value of acceleration just as a number.
The direction of the vector acceleration in this case is opposing the direction of movement, so we can call it rate of deceleration.

The towbar is compressed by the inertia of the caravan, which does not have brakes, I assume.
That is the reason it is a bar and not a rope, cable or chain.

 

[AN630078]

Your calculations for i) and for ii) seem to be corrrect.

As long as you know what is happening and not doing a vectorial addition, you can keep the value of acceleration just as a number.
The direction of the vector acceleration in this case is opposing the direction of movement, so we can call it rate of deceleration.

The towbar is compressed by the inertia of the caravan, which does not have brakes, I assume.
That is the reason it is a bar and not a rope, cable or chain.

Thank you very much for your reply I greatly appreciate it! Thank you I will keep that in mind.
Of course, yes that makes complete sense, I had neglected to think about the effect of the caravan not having brakes. Thank you again.

 

[Lnewqban]

You are welcome