need help in M1

[T.Q]

A car of mass 800 kg pulls a trailer of mass 200 kg along a straight horizontal road using
a light towbar which is parallel to the road. The horizontal resistances to motion of the
car and the trailer have magnitudes 400 N and 200 N respectively. The engine of the car
produces a constant horizontal driving force on the car of magnitude 1200 N. Find

(a) the acceleration of the car and trailer,

(3)

(b) the magnitude of the tension in the towbar.

(3)

The car is moving along the road when the driver sees a hazard ahead. He reduces the force
produced by the engine to zero and applies the brakes. The brakes produce a force on the
car of magnitude F newtons and the car and trailer decelerate. Given that the resistances
to motion are unchanged and the magnitude of the thrust in the towbar is 100 N,

(c) find the value of F.

(7)



i need help in part c with explanation plz

[astarmathsandphysics]

Resultant force on car=100-400-F=ma=800a s0 F=-300-800a (1)
You need the acceleration to find F
Find this by looking at the triler
F=ma
-100-200=200a s0 a=-1.5

sub into (1) F=-300-800*-1.5=900N

[T.Q]

Resultant force on car=100-400-F=m=800a s0 F=-300-800a (1)
You need the acceleration to find F
Find this by looking at the triler
F=ma
100-200=200a s0 a=-0.5

sub into (1) F=-300-800*-0.5=100N

thanx but in the MS the answer is F=900
and this Q FROM JUNE09 EDEXCEL PAPER

[astarmathsandphysics]

Can you post the ms. I have checked and I still get 100N

[T.Q]

For trailer: 200 + 100 = 200f or -200f
f = 1.5 m s-2 (-1.5)
For car: 400 + F – 100 = 800f or -800f
F = 900
(N.B. For both: 400 + 200 + F = 1000f )




this is the answer in the mark scheme

[astarmathsandphysics]

I know my mistake. Few mins

[astarmathsandphysics]

I corrected my answer above

[T.Q]

thanx

[astarmathsandphysics]

Can you post the ms anyway? I dont have it.

[XFlufferzX]

Sir~check out this thread :
https://studentforums.biz/index.php/topic,5492.msg167249.html#msg167249

[astarmathsandphysics]

THANKS

[halosh92]

Resultant force on car=100-400-F=ma=800a s0 F=-300-800a (1)
You need the acceleration to find F
Find this by looking at the triler
F=ma
-100-200=200a s0 a=-1.5

sub into (1) F=-300-800*-1.5=900N

i have a question , why dont we consider the tension in our calculations?
and one more question: wat do they mean by thrust of the towbar? in which direction is it, why is it in a different direction in the second formula??
plz answer ASAP

[M-H]

sirrrrrrr i hav the same ques...??
hw n y do u ignore the tension? plzzzzzzz rply asap!!!

[astarmathsandphysics]

I think the tension is 100. Not at home now.

[halosh92]

I think the tension is 100. Not at home now.

nop thats not the tension