I did the 2001 Past Paper recently and think I've found a mistake in one of Mr Malloy's answers on his website at this link: http://www.webtribe.net/~malloyac/Physics2001solns.html
Q21 has a mass of 18kg being pulled along a frictionless surface from rest by a 4N force at an angle of 26' to the horizontal.
Q21ai asks you to calculate the horizontal force, which is simply Fcos26 = 4cos26' = 3.6N. Fine with that.
Q21aii then asks you to calculate the acceleration along the horizontal surface.
Mr Malloy gives the answer as a = F/m = 4/18 = 0.222m/s2.
However, isn't the answer a = F/m = 3.6/18 = 0.2m/s2 ?
Just a wee bit worried in case there's something really simple I'm missing.
Nobody else references it on this forum, hence my apprehension that Mr M is wrong.
Can someone who knows enlighten me? Anyone have the 2001 past paper to confirm the correct answer? Thanks in advance!
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2001 Q21
Started by JasMcG, Dec 03 2007 01:34 PM
3 replies to this topic
#2
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Posted 04 December 2007 - 03:47 PM
I'm no expert, but it seems that you're right and there's a little mistake in those solutions.
#3
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Posted 04 December 2007 - 07:05 PM
Definitely, you're right.
Acceleration always goes the same direction with its own force. Therefore, the horizontal acceleration has to work out from the horizontal force. There is no point to pick up the horizontal acceleration with the 4N force, they are all different direction
Generally, You are right, well-done!
: .
Acceleration always goes the same direction with its own force. Therefore, the horizontal acceleration has to work out from the horizontal force. There is no point to pick up the horizontal acceleration with the 4N force, they are all different direction
Generally, You are right, well-done!
: .
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