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A level maths help!!!!

Q.In a physics experiment,two balls A and B,of mass 4m and 3m respectively,are travelling towards one another on a straight horizontal track.Both balls are travelling with a speed of 2 ms-1 immediately before they collide.
As a result of the impact,A is brought to rest and the direction of motion of B is reversed.
Modelling the track as smooth and the balls as particles,
a)find the speed of B immediately after the collision
A student notices that after the collision ,B comes to rest 0.2m from A.
B)Show that the coefficient of friction between B and the track is 0.113,correct to 3 decimal places.

This is from Mechanics Solomon Papers
I don't understand how to do no-b.Our teacher first used F=ma and then used F=uR(meow R,formula for maximum frictional force)then wrote ma=uR then tried to find the things.Can anyone explain me what to do exactly?
Original post by Anlasan
Q.In a physics experiment,two balls A and B,of mass 4m and 3m respectively,are travelling towards one another on a straight horizontal track.Both balls are travelling with a speed of 2 ms-1 immediately before they collide.
As a result of the impact,A is brought to rest and the direction of motion of B is reversed.
Modelling the track as smooth and the balls as particles,
a)find the speed of B immediately after the collision
A student notices that after the collision ,B comes to rest 0.2m from A.
B)Show that the coefficient of friction between B and the track is 0.113,correct to 3 decimal places.

This is from Mechanics Solomon Papers
I don't understand how to do no-b.Our teacher first used F=ma and then used F=uR(meow R,formula for maximum frictional force)then wrote ma=uR then tried to find the things.Can anyone explain me what to do exactly?


Due to TSR policies, I can't give you the answer directly.

I think it's important to note that F= ma is not the same as the F in F=uR. I'm relabelling these:
Force = ma and Friction = uR
Force is the resultant force where Force = initial force - Friction

The above should give you enough to figure out the rest.
Reply 2
What did you get for part (a)?
What did you get for the acceleration of B after impact?
Original post by Anlasan
Q.In a physics experiment,two balls A and B,of mass 4m and 3m respectively,are travelling towards one another on a straight horizontal track.Both balls are travelling with a speed of 2 ms-1 immediately before they collide.
As a result of the impact,A is brought to rest and the direction of motion of B is reversed.
Modelling the track as smooth and the balls as particles,
a)find the speed of B immediately after the collision
A student notices that after the collision ,B comes to rest 0.2m from A.
B)Show that the coefficient of friction between B and the track is 0.113,correct to 3 decimal places.

This is from Mechanics Solomon Papers
I don't understand how to do no-b.Our teacher first used F=ma and then used F=uR(meow R,formula for maximum frictional force)then wrote ma=uR then tried to find the things.Can anyone explain me what to do exactly?
Reply 3
Original post by vc94
What did you get for part (a)?
What did you get for the acceleration of B after impact?

I got v=2/3ms-1
And for the acceleration part,I couldn't solve it.In the mark scheme,it is written R=mg
-F=ma but F=uR so a=-uR/m or a=-umg/m so a=-u/g.
Reply 4
Original post by MindMax2000
Due to TSR policies, I can't give you the answer directly.

I think it's important to note that F= ma is not the same as the F in F=uR. I'm relabelling these:
Force = ma and Friction = uR
Force is the resultant force where Force = initial force - Friction

The above should give you enough to figure out the rest.

Okay but how do I get force in the first place?
Reply 5
Original post by Anlasan
Okay but how do I get force in the first place?

Use suvat to get the deceleration over the 0.2 metres, then multiply by mass to get the re tarding force. Then equate that to mu*R, as friction is the re tarding force, and solve as when the object is moving, the (dynamic) frictional force will be at its limiting (static) value.

You could also get the force (or equate directly to muR) using
change in KE = work done
but its effectively the same thing as using suvat. Maybe try both approaches to understand why?
(edited 6 months ago)

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