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Solution to Test Paper 2

From Mechanics

(Difference between revisions)

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Solutions
1 (a)	44.1=219.8t2t=4.944.1=3 sORs=219.832=44.1Hits ground after 3 secondsAG	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=29.4mORv=0+9.83v=29.4	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−2Car is slowing down	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)		B1	(1 mark)	B1: Correct force diagram
4 (b)	100a=200−980sin5a=100200−980sin5=1.15 ms−2	M1A1 M1 A1	(4 marks)	M1: Three term equation of motion A1: Correct equation M1: Rearranging equation. A1: Correct a.
4 (c)	s=05+211.1552=14.4 m	M1 A1 A1	(3 marks)	M1: Using a constant acceleration equation A1: Correct equation A1: Correct distance.
			(8 marks)
5 (a)	v=(6i+4j)+(0.2i−0.4j)t	M1 A1	(2 marks)	M1: Use of v=u+at A1: Correct expression
5 (b)	4−0.4t=0t=10 s	M1 A1 A1	(3 marks)	M1: Using j component equal to zero. A1: Correct equation. A1: Correct time.
5 (c)	r=(6i+4j)30+21(0.2i−0.4j)302=270i−60jr=2702+602=277 m	M1 A1 M1 A1	(4 marks)	M1: Using r=ut+21at2 A1: Correct position vector. M1: Finding distance. A1: Correct distance
			(8 marks)