PHYS1131 HIGHER PHYSICS 1A SOLUTIONS Homework Problem Set 1. To calculate the time taken, we substitute d=10 m into the above equation.

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1 PHYS1131 HIGHER PHYSICS 1A SOLUTIONS Homework Problem Set 1 Q1. We use the equations of uniform accelerated motion with zero initial displacement and velocity. y 1 = " 1 gt where t is measured in seconds. Then: y = " 1 gt' = " 1 g(t "1) d = y 1 " y = 1 gt " 1 g(t "1) = gt " g / where t' = t-1 To calculate the time taken, we substitute d=10 m into the above equation. t = d + g / g =14.9/9.8 =1.5s Q. Take y=0 as the height of lift s floor when the bolt starts to drop. Again, we use the equations of accelerated motion. y = y o + v yo t + 1 a y t y lift = 0 + t t y bolt = 3 + t " 4.9t S.I.units The bolt hits the floor when y lift = y bolt Solving these simultaneously, 3 + t " 4.9t = t t # 3 = 5.65t # t = 0.73s At this time, the lift is at a position: y lift =1.86m "d fallen = ( 3#1.86)m =1.1m

2 Q3. a) x = k ( 1 e kt ) x k b) v = dx dt c) a = dv dt = e kt = k e kt = kv t d) (c) implies a resistive force proportional to v. This force becomes weaker with as speed becomes lower in such a way that the object never stops. The distance travelled is finite, however, because the integral of v.dt is finite. Past exam question: y h T T i) Depending on the choice of origin, the graphs might look like these. The algebra below is for the upper. ii) y = y o + v yo t + 1 a y t falling stone T rising sound = h gt hits the bottom when y = 0, so 0 = h 1 gt 1 T 1 (T + T 1 ) t " T 1 = h g T 1 = h g iii) h = 78 m " T 1 = 4.0 s.

3 y T T taken, so T = h/v s = 0.3 s. v) speed = distance travelled/time falling stone rising sound vi) T = T 1 + T = 3.99 s s "# = 4. s. ( significant figures) T -h T 1 (T + T 1 ) t vii) She says 0 = h 1 gt 1 " h = 1 gt 1 $ So she calculates 1 gt. h $ 1 (10ms - )(4 s) = 80 m. viii) Her time estimate is 4.0 ± 0.5 s, an error of 13%. Neglecting the time for the sound to travel (6%) and taking 9.8 $ 10 (%) are small errors by comparison. <She is lucky to have worked out an answer so close to the precise one.> (In the marking scheme, any reasonable comment about the accuracy earns two marks.) Q4. r 1 = ( 100m) i r = "( 300m) j r 3 = "( 150m)cos30 o i" ( 150m)sin 30 o j = " 130m " 75m ( ) i ( ) j " r 4 = "( 00m)cos60 0 i+ ( 00m)sin60 0 j r = r 1 + r + r 3 + r 4 = " 130m ( ) i r = r r = ( 130m) + "0m # = tan "1 $ r ' y & % r ) = 37o x ( " 0m ( ) j ( ) = 40m = ("100m,173m)

4 Q5. r = a " b + c =11.0 i+ 5.0 j" 7.0 k r = r x + r y + r z =14.0 " z = cos #1 ( r z r ) =10 0 Q6. With respect to the ground, let the wind have velocity and the person (initially) have v p. Let the person s first frame have one dash, and the relative velocity of the wind be '. Let the second frame have two dashes, and the relative velocity of the wind be ". We have, in the dash frame, = v p + ', which is shown below at left. We have, in the double dash frame, = v p + ", which is shown below at right. v p v p N ' v p ' v p 45 " On the right, we are given that the angle at top right is 45, but because the two lengths on the top side are each v p, the angle at top left must be 45 too, so the wind must come from NW and = v p /cos45 = 5.7 kph. (It is also possible to do this by vector algebra.) Q7. The velocity of the rower with respect to the bank is v rb. The velocity of the rower with respect to the water is v rw. The velocity of the water with respect to the bank is b. So a) v rb = v rw + b = v rw sin % î + v rw cos % j + b î = [(5 sin %)î + cos % j ] m.s 1. At this point, some students may choose to use numerical values for the rest of the problem, which would be acceptable in a test. This makes the equations a little shorter. It has two disadvantages: first, one cannot readily do checks for dimensions and limits. Second, an algebraic solution is a solution for all cases, and is also the starting point for further problems. b) r rb = v rb.t + 0 = t [ v rw sin % î + v rw cos % j + b î ] (1) = t [(5 sin %)î + cos % j ] m.s 1. c) At the opposite bank, y = 40 m = w. Substituting for the y component in (1) gives: y = t [v rw cos %]

5 t = w v rw cos % () t is a minimum when cos % is a maximum, ie when % = 0, so here % = 0 and t = 40m m.s-1 = 0 s. Distance D from B on the other bank = the final x component of r rb. From (1), D = t [ v rw sin % + b ] (3) so, in this case, D = 0 s [0 + 5 m.s 1 ] = 100 m. d) To remove t, from (3), substitute from (): D = w( v rw sin % + b ) v rw cos % (4) for which we must find a minimum. dd d% = w v rw cos % v rw cos % + w( v rw sin % + b ) v rw cos % sin % = 0 multiply all terms by v rwcos % v rw w gives cos % sin % + b v rw sin % = 0 b v rw sin % = 1 % = sin -1 v rw b (= 3.6 ) = 0 (to one significant figure) Substitute (the precise value) in () and (3) or (4) gives t = 0 s and D = 90 m. Q8. " v = (100 i# 75 j) # (15 i+ 5 j)m.s #1 = #5 i#100 jm.s #1 a av = " v "t = 1 3 (#5,#100)ms# = 8.3i# 33 j ms #1 (SI units) Q9. (a) r (t) = ˆ i + 4t ˆ j + tk ˆ v (t) = d r dt = 8tˆ j + k ˆ a (t) = d v dt = 8 ˆ j (SI units) so the acceleration is constant in the y-direction.

6 (b) x =1; y = 4t " $ # z = t % $ y = 4z so the particle moves along a parabolic path in the plane x=1. Q10. Use subscript 0 for initial and f for final. We are given the acceleration (a y = g) and the time of flight t f. We are also given the initial speed and angle %, which gives us the initial velocity components: v x0 = cos % and v y0 = sin % a) what is final height y f = h? We know v y0, a y and t f so use y = y0 + vy0t + 1 a yt h = y f = 0 + v0sin% t f + 1 a yt =... = 51.8 m b) final speed is v f = v xf + v yf a x = 0 so v xf = v x0 = cos % a y = g so v yf = v y0 + a y t f. v f = v xf + v yf = ( cos %) + ( sin% gt f ) = 7.4 m.s -1 c) Max height y m = H occurs when v y = 0 v y v y0 = a y (y y 0 ) applied here gives 0 ( sin%) = gh H = ( sin%) /g = 67.5 m Q11. V 0 y Vertical motion : y = 1 gt The ball falls to level of nth step in time t n = y / g = 0.0 n s. Horizontal motion: x n = t n. N 1 3 t n (s) x n (m) By examination of the above data, the ball hits the second step (x < x 5 cm) x < * 5cm) < x 3

7 Q1 & '( Remember: y = y0 + vy0 + 1 at ; vy = vy0 + ayt ) *+ a = 1.0 * m.s Vertical motion: v = 0 + at (1) Horizontal motion: x = 0 + vot + 0 y = 0 + 0*t + 1 at () Time T to pass through plates: T = L/vx =.0 ns. a) substitute T in () to get.0 mm b) vx = v0 vy(t) = at v = vx + vx so v(t) = v0 + a T = 1.0*10 7 m.s 1 % = tan 1v y vx so %(T) = tan 1 & '( at v0) *+ = 11. i.e. vf = 1.0*10 7 m.s 1 at 11 below horizontal. (~ 3% of c, so = so relativity neglected for sig figs.) Q13. (a) a = v r " v = gr =19.8m/ s (b) It will appear weightless, but is not weightless. See discussion 'weightless' on the HSC Physics FAQ at Q14. # & r = r cos"t i + r sin "t j % $ ~ ( ~ ' v = d # & r dt = r" )sin "t i + cos"t j % $ ~ ( ~ ' a = d # & v dt = )" r cos"t i + sin"t j % $ ~ ( = )" r ~ '

8 Q15. v x = dx dt ="R(cos"t +1) v y = dy dt = #"Rsin"t a x = #" Rsin "t a y = #" Rcos "t,t v x v y a x a y 0,R 0 0 -, R , R See rolling on Physclips. Q16. (a) x = ( cos")t y = h + ( sin")t #1/gt (b) Eliminating t from the above equations, we get: x y = h + sin"( cos" ) # 1 g( x cos" ) $ y = h + x tan" # gx sec " Using y=0, x=r, we obtain: 0 = h + R tan" # gr sec " (1) as required.. (c) Differentiating the above expression with respect to theta, the result is: 0 = (tan" # gr sec ") dr d" + Rsec "(1# gr tan") dr d" = 0 $1# gr tan" = 0 $ R = gtan"

9 (d) Substituting this value for R into (1): 0 = h + v 0 g " g 4 v v sec # $ 0 0 g tan # (e) = h + v 0 g " gsin # %sin # = gh / 0.5 sin " = #.1/14 = $" = $ R =.0 m (f) 45 o 4 o When h = 0 the maximum range is indeed given for % = 45 o (exercise for the student]. However for h > 0 then the projectile has an additional time to travel before it hits the ground, and the extra distance travelled will be maximised if there is additional horizontal speed, i.e. this occurs if % < 45 o. Past exam question: i) There is no air resistance. Neither the bird's nor the grape s, horizontal velocity changes, so, if they have the same horizontal velocity, they always have it. If they ever have the same horizontal position, they must always have it. So you must throw it when the bird is directly overhead. ii) From (i), vxg = v0 cos % = vb. (a) Vertical motion under gravity, measured from y = 0 at the position of the head: vy = vy0 + ay At y = h, vy = 0, so 0 = v0 sin % gh v0 sin% = gh (b) (b)/(a). tan % = (a). v0 = vb cos % gh vb = 5 m.s-1 cos 63 so % = tan -1 = 11 ms -1. gh vb = tan -1 *9.8ms - *5m 5 m.s -1 = 63 iii) Air resistance would slow the grape during flight. The grape would have greater horizontal velocity until the end of its flight, so it would cover the distance from you to bird faster than the bird would, so you would throw it after it passed overhead.