محاضرة 14 فيزياء عامة (1) ميكانيكا نيوتن .. تمارين محلولة على الفيزياء والقياس
Introduction: Physics and Measurements
[1] Show that the expression x=vt+1/2at2 is dimensionally correct, where x is a coordinate and has units of length, v is velocity, a is acceleration, and t is time.
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نقوم بإيجاد وحدة الطرف الأيمن ونساويها بوحدة الطرف الأيسر من المعادلة وفي حالة تساوي الأبعاد (الوحدات) لطرفي المعادلة فإن المعادلة تكون صحيحة وإلا فإنها خاطئة.
الإجابة: المعادلة صحيحة |
[2] Which of the equations below are dimensionally correct?
(a) v = vo+at (b) y = (2m)cos(kx), where k = 2 m-1. اعلانات جوجل
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بعد حل المعادلة بنفس الطريقة السابقة نجد أن:المعادلة (a) صحيحة المعادلة (b) صحيحة |
[3] Show that the equation v2 = vo2 + 2ax is dimensionally correct, where v and vo represent velocities, a is acceleration and x is a distance. |
The L.H.S has a unit of (m/s)2
The R.H.S has a unit of (m/s)2 + (m/s2) m = (m/s)2 L.H.S = R.H.S (the equation is correct) |
[4] The period T of simple pendulum is measured in time units and given by
where l is the length of the pendulum and g is the acceleration due to gravity. Show that the equation is dimensionally correct. |
بعد تحديد ابعاد الطرف الأيمن وابعاد الطرف الأيسر نحصل على أن النتيجة التساوي وهذا يعني أن المعادلة صحيحة |
[5] Suppose that the displacement of a particle is related to time according to the expression s = ct 3. What are the dimensions of the constant c? |
حتى تكون المعادلة صحيحة فإن وحدة الطرف الأيمن من المعادلة يجب أن يكون مساوي لوحدة الطرف الأيسر وهو وحدة طول (m) فإن وحدة الثابت c هي m/s3 |
[6] Two points in the xy plane have Cartesian coordinates (2, -4) and (-3, 3), where the units are in m. Determine (a) the distance between these points and (b) their polar coordinates. |
[7] The polar coordinates of a point are r = 5.5m and q = 240o. What are the cartisian coordinates of this point? |
x = r cos q = 5.5×cos 240o = -2.75 m
y = r sin q = 5.5×sin 240o = -4.76 m
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[8] A point in the xy plane has cartesian coordinates (-3.00, 5.00) m. What are the polar coordinates of this point? |
المراد من السؤال هو التحويل من الاحداثيات الكارتيزية إلى القطبية.
θ = 121o with respect to the positive x-axis
(-3,5)m = (5.8m, 121o) |
[9] Two points in a plane have polar coordinates (2.5m, 30o) and (3.8, 120o). Determine (a) the cartisian coordinates of these points and (b) the distance between them. |
The Cartesian coordinate for the point (2.5m, 30o) x = 2.16 m & y = 1.25 m A = 2.16i + 1.25j |
The Cartesian coordinate for the point (3.8m, 120o)
x = -1.9 m & y = 3.3 m
B = -1.9i + 3.3j
(b) The distance between the two points is the magnitude of the displacement vector.
B – A = 4.06i + 2.05j
|B-A| = 4.5 m
[10] A point is located in polar coordinate system by the coordinates r = 2.5m and q =35o. Find the x and ycoordinates of this point, assuming the two coordinate system have the same origin. |
r = 2.5 , θ = 35o
x = r cos 35 = 2 y = r sin 35 = 1.4 |
[11] Vector A is 3.00 units in length and points along the positive x axis. Vector B is 4.00 units in length and points along the negative y axis. Use graphical methods to find the magnitude and direction of the vectors (a) A + B, (b) A – B. |
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[12] A vector has x component of -25 units and a y component of 40 units. Find the magnitude and direction of this vector. |
نقوم برسم المتجه من معطيات السؤال
A = -25i + 40j
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[13] Find the magnitude and direction of the resultant of three displacements having components (3,2) m, (-5, 3) m and (6, 1) m. |
نحول كل نقطة من النقاط الثلاثة في السؤال إلى الصورة المتجهة كما يلي:
A = 3i + 2j B = -5i + 3j C = 6i + j نوجد المحصلة بالجمع الإتجاهي A + B + C = 4i + 6j |
[14] Two vector are given by A = 6i –4j and B = -2i+5j. Calculate (a) A+B, (b) A-B, |A+B|, (d) |A-B|, (e) the direction of A+B and A-B. |
أنظر المثال رقم 8 في الكتاب المقرر صفحة رقم 17 |
[15] Obtain expressions for the position vectors with polar coordinates (a) 12.8m, 150o; (b) 3.3cm, 60 o; (c) 22cm, 215 o. |
(a) 12.8m, 150o
x = r cos θ = 12.8 cos 150 = -11.1m y = r sin θ = 12.8 sin 150 = -17.5 m
A = -11.1i – 17.5j استخدم نفس الطريقة لباقي النقاط لإيجاد متجه الموضع. |
[17] A vector A has a magnitude of 35 units and makes an angle of 37o with the positive x axis. Describe (a) a vector B that is in the direction opposite A and is one fifth the size of A, and (b) a vector C that when added to A will produce a vector twice as long as A pointing in the negative y direction. |
(a) the vector B is equal to -1/5 A
A = Axi + Ayj A = 28i + 21j B = -5.6i – 4.2j (b) The vector C + A = R R = twice as long as A pointing in the negative y direction Magnitude of A is equal to 35 R = 2*35 = 70 units R = -70j C = R – A therefore C = -70j – 28i – 21j = -28i -91j |
[18] Find the magnitude and direction of a displacement vector having x and y components of -5m and 3m, respectively. |
In unit vector notation
A = -5i + 3J The magnitude is The direction with respect to the negative x-axis |
[19] Three vectors are given by A=6i, B=9j, and C=(-3i+4j). (a) Find the magnitude and direction of the resultant vector. (b) What vector must be added to these three to make the resultant vector zero? |
A=6i, B=9j C=(-3i+4j) |
The resultant vector is A + B + C = 3i + 13j
The Magnitude of the resultant vector is 13.34 units
The direction is 77o with respect to the positive x-axis
(b) The vector must be added to these three to make the resultant vector zero is
-3i – 13j
[20] A particle moves from a point in the xy plane having cartesian coordinates (-3.00, -5.00) m to a point with coordinates (-1.00, 8.00) m. (a) Write vector expressions for the position vectors in unit-vector form for these two points. (b) What is the displacement vector? |
The vector position for the first point (-3,-5)m is
A = -3i -5j The vector position for the first point (-1,8)m is B = -i + 8j (b) The displacement vector is B – A = 2i +3j |
[21] Two vectors are given by A= 4i+3j and B= -i+3j. Find (a) A.B and (b) the angle between A and B. |
(a)
A.B = AxBx+AyBy A.B = -4 + 9 = 5 units (b) cos θ = A.B/AB = 1/3.16 θ = 71.6o |
[22] A vector is given by A= -2i+3j. Find (a) the magnitude of A and (b) the angle that A makes with the positive y axis. |
(a) the magnitude of A = 3.6 unit
(b) cos θ = 3/3.6 θ = 33.5o |
[23] Vector A has a magnitude of 5 units, and B has a magnitude of 9 units. The two vectors make an angle of 50o with each other. Find A.B |
A.B = A B cos θ
A.B = 5 x 9 cos 50o = 28.9 unit |
[24] For the three vectors A=3i+j-k, B= -i+2j+5k, and C= 2j-3k, find C.(A-B) |
A – B = 4i -j -6k C = 2j – 3k |
C.(A-B) = 0 -2 + 16 = 14 unit
[25] The scalar product of vectors A and B is 6 units. The magnitude of each vector is 4 units. Find the angle between the vectors. |
A.B = 6 units
A = B = 4 units cos θ = 6/16 θ = 67.9o |