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Solution to Test Paper 2

Revision as of 15:59, 19 January 2011 by Ian (Talk | contribs)

Solutions
1 (a)	44 .1=219.8t2t=4.944.1=3 sORs=219.832=44.1AGHits ground after 3 seconds	M1 A1 A1 (M1) (A1) (A1)	(3 marks)	M1: Use of constant acceleration equation with v=0 A1: Correct equation A1: Correct s
1 (b)	v2=02+29.844.1v=864.36=29 .4 ms−1ORv=0+9.83v=29.4 ms−1	M1 A1 A1	(3 marks)	M1: Use of constant acceleration equation with v=0 A1: Correct equation. A1: Correct v.
1 (c)	Air resistance would slow the ball down.	B1	(1 mark)	B1: Sensible statement about air resistance.
			(7 marks)
2 (a)		B1	(1 mark)	B1: Correct horizontal forces. Ignore any vertical forces.
2 (b)	P=900N	B1	(1 mark)	B1: Correct value for P.
2 (c)	P−900=20001.2P=2400+900=3300N	M1 A1 A1	(1 mark)	M1: Three term equation of motion A1: Correct equation A1: Correct P.
2 (d)	800−900=2000aa=2000−100=−0.05 ms−2Carisslowingdown	M1 A1 A1 A1	(4 marks)	M1: Three term equation of motion A1: Correct equation A1: Correct a A1: Correct statement
			(9 marks)
3 (a)	R=209.8=196 N	M1 A1	(2 marks)	M1: Use of R=mg A1: Correct R.
3 (b)	F=0.4196=78.4 N	M1 A1	(2 Marks)	M1: Use of F=R A1: Correct F.
3 (c)	100−78.4=20aa=20100−78.4=1.08 ms−2	M1 A1 A1	(3 marks)	M1: Three term equation of motion A1: Correct equation A1: Correct a.
			(8 marks)
4 (a)		3	4	5
4 (b)	2	3	4	5
4 (c)	2	3	4	5
1	2	3	4	5
5 (a)	2	3	4	5
5 (b)	2	3	4	5
5 (c)	2	3	4	5