Answers to Coursebook questions Chapter 2.10

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1 Camride Physis for the IB Diploma Answers to Courseook questions Chapter. 1 a y = OP = 1 t = 0.05 m = 0.0 = 00 m s 1 0. The time to fall to the floor is iven y y = 1 t t = y = 1.3 = 0.51 s. The horizontal distane travelled is therefore x = t = = m. 3 a The time to fall to the floor is iven y y = 1 t t = y = 3 = s. The speed an e found y findin the omponents of the veloity: = 5.0 m s 1, v y = t = 7.75 m s 1. Then v = = 9. m s 1. Alternatively, we an use onservation of enery to find 1 mu + mh = 1 mv v = u + h = = 9. m s 1. 4 a Use y = 1 t t = y = 0 =.0 s. After 1 s, = 8.0 m s 1, v y = t = m s 1. The speed is thus v = = m s 1. The anle is θ = tan 1 v y = tan 1 ( 1.5) = 51 o, i.e. 51 o elow the horizontal. d = 8.0 m s 1, v y = t = 0 m s 1 and so the speed is v = = 1.5 m s 1 at an anle θ = tan 1 v y = tan 1 (.5) = 68 o. 5 The times to hit the round are found from y = 1 t t = y = = s and = 1.65 s. The ojets are thus separated y 4.0 ( ) = m when they land. Copyriht Camride University Press 011. All rihts reserved. Pae 1 of 6

2 Camride Physis for the IB Diploma 6 The horizontal distane travelled y the ojet fallin from 8.0 m is (see Q5) x = t = = 5.06 m. Thus the speed of the other ojet must e = = m s The pakae will fall in a time = 6.35 s and so the horizontal distane travelled will e x = t = = m. 8 The reatest rane is otained when the launh is at 45 o. Let the speed of launh e v. Then y = vt sin 45 o 1 t. y = 0 when 0 = vt sin 45 o 1 t vsin 45o t =. This is the time the disus is in the air. Then the rane is vsin L = vt os 45 o 45o = v os 45 o = v sin 45 o os 45 o v = = v Hene v = L. The maximum vertial heiht is found from onservation of enery: 1 mv = mh h = v = L = L The keys will fall in a time = 1.65 s and so the horizontal distane travelled will e x = t = = 7.6 8m. The omponents of veloity are: a = vos40 o = 0os40 o =15.3 m s 1 and so the raph is a horizontal straiht line and = vos40 o t = (1.9 t)m s 1 and so the raph is a straiht line with neative slope as shown in answers in textook. The aeleration is onstant so raph is a horizontal straiht line. 11 v y = vsin 40 o t. At the hihest point this omponent is zero and so t = Then, from y = vt sin 40 o 1 t we find vsin 40o = 4.0 sin 40o = 0.57 s. y = sin 40 o = m. Copyriht Camride University Press 011. All rihts reserved. Pae of 6

3 Camride Physis for the IB Diploma 1 The veloity omponents after 1.0 s are = vos35 o = 6.0os35 o = m s 1 and v y = vsin 35 o t = 6.0sin 35 o 1.0 = m s 1. Thus v = ( 6.559) = m s 1. The diretion of the veloity vetor is at θ = tan v 1 y = tan 1 ( ) = 53o. 13 a The horizontal displaement is iven y x = 0t os50 o = 1.6t whose raph is a straiht line. The vertial displaement is y = 0t sin50 o 1 t = 15.3t 5t whose raph is a onave up paraola. 14 a The potential enery is iven y E P = my = 40(0t sin 30 o 1 t ) = 400t 00t whose raph ives the paraola in the answers on pae 797 in Physis for the IB Diploma. The kineti enery is iven y E K = E T E P = 1 mu (400t 00t ) = t + 00t whose raph ives the paraola in the answers on pae 797 in Physis for the IB Diploma. 15 In time t the monkey will fall a vertial distane y = 1 t ut so will the ullet and hene the ullet will hit the monkey. 16 a The all overs a horizontal distane of 0 m in.0 s and so the horizontal veloity omponent is 0.0 = m s 1. An arrow direted vertially down. The horizontal distane etween the dots will stay the same. The all will stay in the air for a shorter time: from y = 1 t we et 0 = 1 0t t = 1.41 s as opposed to.0 s and so the rane will e shorter, x = t = 1.41 = m. The vertial distane etween the dots will e reater, hene the raph in the answers on pae 798 in Physis for the IB Diploma. Copyriht Camride University Press 011. All rihts reserved. Pae 3 of 6

4 Camride Physis for the IB Diploma 17 a The all overs a horizontal distane of 60 m in.0 s and so the horizontal veloity omponent is = 60.0 = 30 m s 1. The all lims to a heiht of m in 1.0 s and so from y + v y t we have u y = 0 m s 1. d e The anle of launh is θ = tan 1 u y = tan 1 ( 0 30 ) = 34o. The vertial omponent of veloity eomes zero at 1.0 s and so v y t 0 = = 0 m s. The veloity is horizontal to the riht and the aeleration is vertially down. With = 40 m s, the all will stay in the air for half the time and so will have half the rane. The maximum heiht is reahed in 0.50 s and is y t 1 t = = 5.0 m, i.e. half as reat as efore. This leads to the raph in the answers on pae 798 in Physis for the IB Diploma. 18 To answer this question we must have formulas for the rane and the maximum heiht. In eneral we have that: x = t and y t 1 t. The time to et to the top is found from 0 t t and so the time to over the rane is doule this, i.e. u y. u y Then x = t 30 = and y t 1 t u y 1 1 u y Simplifyin these two equations we et: u y = 150 (u y ) = 40. The seond implies that u y = 15.5 m s 1 and the first that = 9.7 m s 1. The speed is thus v = = 18 m s 1 at θ = tan 1 u y = tan 1 ( ) = 58o to the horizontal. Copyriht Camride University Press 011. All rihts reserved. Pae 4 of 6

5 Camride Physis for the IB Diploma 19 a The initial veloity omponents are: = 0.0os48 o =13.38 m s 1 and u y = 0.0sin 48 o = m s 1. The all hits the sea when the vertial displaement is y = 60.0 m. Thus y t 1 t 60.0 = 14.86t 5.00t. Solvin for the positive root we find t = 5.6 s. Hene = = m s 1 and v y t = = m s 1. The speed at impat is thus v = ( 37.74) = 40 m s 1 at θ = tan 1 v y = tan 1 ( ) = 70o to the horizontal. Some of the kineti enery of the all will e onverted into thermal enery and so the speed at impat will e less. The horizontal omponent of veloity will derease in the ourse of the motion and will tend to o to zero, ut the vertial omponent will never eome zero (after reahin the maximum heiht). This means that the anle of impat will e steeper. 0 The omponents of displaement are iven y x = vt osθ and y = vt sinθ 1 t. Clearly x = d osφ and so d osφ = vt osθ t = d osφ v 0 osθ. d osφ Then y = v 0 v 0 osθ sinθ 1 d osφ v 0 osθ But also y = d sinφ and so Solvin for d we find d = v 0 osθ (osφ sinθ sinφ osθ) os φ = v 0 osθ sin(θ φ) os φ. d osφ d sinφ = v 0 v 0 osθ sinθ 1 d osφ v 0 osθ osφ sinθ sinφ = 1 osθ d os φ v 0 os θ 1 The net fore on the all is the omponent of its weiht down the inline, i.e. msinθ. The aeleration of the all is thus sinθ = 5.0 m s. The motion of the all is thus idential to that of a all projeted into the air in a uniform ravitational field of strenth 5.0 m s and so is the same as projetile motion. In the asene of resistane fores the path will e paraoli. We then have = 5.0os5 o = 4.53 m s 1 and u y = 5.0sin5 o =.11 m s 1. The all reahes the maximum heiht on the inline when 0 = t t = 0.4 s and so y = (0.4) = 0.45 m. Copyriht Camride University Press 011. All rihts reserved. Pae 5 of 6

6 Camride Physis for the IB Diploma After 1.0 s the veloity omponents are: = and v y. Then tan0 o. The maximum heiht is reahed at a time iven y: 0 t t and so the maximum heiht is iven y: y = 0 t 1 t u y 5(u y ). This simplifies to: 0 = (u y ) 0 and so u y = 0 m s 1. From tan0 o we then find tan0 o = m s 1. The launh speed is then = 34 m s 1. (The anle of launh is θ = tan v 1 y 0 = tan 1 ( 7.47 ) = 36o.) Copyriht Camride University Press 011. All rihts reserved. Pae 6 of 6