S 1 a) Speed is not constant because the vertical component keeps changing b ) Acceleration is constant, always 9.81 m s -2 , always directed downward (regardless of whether the projectile is moving upward or downward) c) Horizontal component of velocity is constant, because acceleration takes place only in the vertical direction. d ) Vertical component of velocity is always changing due to gravity.

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S 2 50 m s -1 300 m a) c ) b)

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S3 (A)

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S4 (c)

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S5 (A)

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S6

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S7

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Immediately after the collision, ball’s speed is same as its speed just before the collision, but directed horizontally. D1 So it is a projectile launched horizontally. The horizontal component of velocity for a projectile is constant. So the horizontal spacing between dots should be constant. The only difference between A and D is the horizontal velocity. D is wrong because it shows the horizontal velocity to be less than the vertical speed just before collision.

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D2 Motion of the ball is a projectile motion with initial velocity 10 m s -1 directed horizontally. So horizontal spacing should be constant. Vertical spacing should become larger and larger

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D3 That they all reached the same maximum vertical height tells us that they were all launched with the same initial vertical velocity component, and had the same time of flight. Since the time of flight is the same, the one with the longest horizontal range had the largest horizontal velocity component, and also the largest initial speed. M aximum vertical height only depends on the initial vertical component of velocity. a=b=c a =b=c a<b<c a<b<c

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D4 (D) Horizontal component of velocity is constant. Since horizontal displacement is doubled, time of flight is also doubled. In doubling the time, the vertical displacement will be quadrupled.

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D5 (B) For the largest £, v x should be as small as possible, v y should be as large as possible.

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D6

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D7 bi) The horizontal spacing remains constant, implying the horizontal component of velocity is constant, implying frictional force is negligible, bii ) Use the 7 th dot for better accuracy. 29.25/7 = 4.17 m s -1 biii ) 40 m biv1) (s=ut+1/2at 2 ) 40 = 0 + 1/2 a 7 2 a =1.63 m s -2 . biv2) mg €h = (0.050)(1.63)(40)=3.26 J bv ) gain in KE = loss in GPE KEf – KEi = 3.26 KEf = 3.26 + 1/2 (0.050) (4.17) 2 KEf = 3.69 J c ) KEf = 1/2 m v 2 3.69 = 1/2 (0.050) v 2 v = 12.15 m s -1 cos £ = v x /v cos £ = 4.17/12.15 £=69.9°

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D7 di) Just copy the graph dii2) The horizontal spacing is exactly the same as the red line. The vertical spacing is a fixed proportion of the red line. dii1) The horizontal spacing becomes less and less, until the horizontal component of velocity reaches zero. The vertical spacing increases at a decreasing rate until it becomes constant when terminal velocity is reached.

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D8 At t=0 . The horizontal distance between the balls remain constant. The vertical distance keeps on increasing. So they were closest right at the start. b ) Yes. Their horizontal velocity is always the same. The first ball’s vertical velocity is always faster than the second. c) 1 second lah . d ) No. The horizontal velocity does not affect the vertical motion in any way.

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D9 Careful. Because we chose upward to be positive, it should be – s y = s x

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D10 ii) These portions of the two trajectories are exactly the same. So the distance between the points of impact is equal to the “horizontal range” of Q.

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C1

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