Magnetism #1 - Questions and Answers

Question 1 Starting with an initial speed of 5.00 m/s at a height of 0.300 m, a 1.50 kg ball swings downward and strikes a 4.60kg ballthat is at rest, as the drawing shows. a. using the principle of conservation of mechanicalenergy,find the speed of the 1.50 kg ball just before impact b. assuming that the collision is elastic, find the velocities(magnitude and direction ) of both balls just after thecollision c. how high does each abll swing after the collision, ignoringair resistance? Answer Let m = 1.5kg, vl the initial speed of m, v2 the speed of mbefore impact, v3 speed after impact By conservation of energy 1 -m 2" +mxgxh= 1 × v1² + m × 9 × h = = × m × v before impact and 2 v22 = v12 + 2 × g× h = 25+2×9.8 x 3 = 30.88 and v2 = 5.56m/s Let M = 1.5, V the speed of M after impact and v3 the speed ofm after impact Since the collision is elastic 1 1 2 xmx 2 2 x m x v3² + 1 × MX V2 2 Rewrite this as m imes (v22 – v3²) = M × V2 (1) By conservation of momentum m x v2 = mx v3 + MXV Rewrite this as m × (v2 – v3) = M × V (II) Divide equation (1) by equation (II) and get v2 + v3 = V Substitute V in equation (II) and m x v2 m x v3 = M x v2 + Mx v3 Solve for v3 and get v3 = v2 x m-M It'sworth noting that if M = m then v3 = 0 M+m 03 = -5.56 × = -2.83 m/s and V = 5.56m/s -2.83m/s = 2.73m/s 3.1 6.1 1 Now you can use × m x v3 = m xgx hand = × M × V2 = xg × H 2 To solve for the height that each ballrises. Question 2 A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q. A) What is the total surface charge q on the interior surface of the conductor (i.s., on the wall of the cavity)? C) What is the magnitude Es of the electric field inside the cavity as a function of the distance from the point charge? Let k, as usual, denote 1 4περ D)Will the total charge on the surface of the cavity change? Will the totaly charge on the exterior of the conductor change? Will the charge on the electric field with in the cavity change? Will the charge on the electric field outside the cavity change? Answer 1 (A) From the property of conduction, a charge of q with opposite sign will induce on the inner surface of the cavity. An equal amount of charge q adds to the exterior surface of cavity. That is qint = -9 and exterior = +q 1 (C) From Gauss's law, the electric flux through any Gaussian surface is equal to ε the enclosed charge. Forr < R (radius of the spherical cavity), the enclosed charge is lenclosed = q. The electric field is derived as follows: Ε(4πν²) = 2 times εο 1 9 E = 2 Απε, να (D) No, the total charge on the surface of the cavity doesn't change. Yes, from the property of induction, the charge on the exterior surface changes by an amount exterior = +q. Yes, the charge on the electric field within the cavity changes Yes, the charge on the electric field outside the cavity changes Answer 2 Explanation: The spherical cavity has a neutral charge. This means that this spherical cavity won't affect the field inside. Therefore the field generated by the charge q+ is the same with or without the spherical cavity. The total charge of the system can be calculated considering charge conservation: Qtot0=Qtotf q(+)=q(+)+qin+qex qin=-qex and qin+qex=0 Using a Gaussian sphear centered in the charge q(+), we can calculate the field knowing that its direction would be (r) If we apply the gaussian sphear between the interior and the exterior radius of the conducting hollow sphear, knowing that the field should be 0: $Eds = fods = 0 = 9(+) + qint = 0 9(+) = -qint 9(+)+qint QIN 60 If we apply the gaussian sphear outside the exterior radius of the conducting hollow sphear, we obtain the field required in c). Question 3 In a truck-loading station at a post office, a small 0.200-kg package is released from rest a point A on a track that is one quarter of a circle with radius 1.60 m. The size of the packageis much less than 1.60m, so the package can be treated at aparticle. It slides down the track and reaches point B with a speed of 4.80 m/s. From point B, it slides on a level surface a distanceof 3.00 m to point C, where is comes to rest. (a) What is the coefficient of kinetic friction on the horizontal surface? (b) How much work is done on the package by friction as it slides down the circular arc from A to B? Answer 1 a.) The kinetic energy of the package at point determines how much work is done by friction. 1 µmgL = =mv 2 Then solve for με (4.8) 2 = 0.392 2gL 2(9.8) (3) ν B με = b) Wother KB-UA 1 Wother = =mv² - mgh 2 ==(0.2) (4.8) 2 – (0.2) (9.8) (1.6) 1 2 0.832 J Answer 2 a) Kinetic friction on horizontal surface is equal to friction force µNd cos(0) = m(v – v) N=mg Given m = 0.200kg; d = 3.00m; v₁ = 4.80m/s; v2 = 0 Therefore μ(0.200kg) × 9.8 × 3 cos 180° = (0.200) × (02 – 4.802) Step 2 b) Total work= work done by gravity + work done by friction Total work = kinetic energy WT = m(v - и²) Wg+W₁ = m(v – u²) W₁ = m(v – u²) – mgR = 0.200((4.80)2 – 0)) – (0.200)(9.8)(1.60) Wf = -0.832J (work done against moving body, hence -ive sign) Answer 3 Step 1 Given: The mass is 2 kg The radius is 1.60 m The package's size is 1.6 m The speed is given as 4.80 m/s The distance is 3 m Step 2 a) At point B, kinetic energy is equal to the friction's work done. The formula is denoted as, mv = μεmg × Bc με = 4.82 UB 2gxBc 2×9.81×3 με = 0.39 Step 3 b) The sum of kinetic energy at point B and work done by the friction is equal to the Potential energy at point A. The formula is denoted as. mgR = WFAB + mv WFAB = m(gR-03) On substituting the values, WFAR = 0.2(9.81 × 1.60 – 0.5 × 4.82) WF = 0.835J FAB Step 4 Solution: a) At the horizontal surface, the coefficient of friction is 0.39 b) From A to B, the friction's work done is 0.835 J Question 4 A constant potential difference of 12V is maintained between the terminals of a 0.25 micro- F parallel plate air capacitor. A. A sheet of Mylar is inserted between the plates of the capacitor completely filling the space between the plates. When this is done how much additional charge flows into the positive plate of the capacitor? (K for mylar = 3.1) B. What is the total induced charge on either face of the mylar sheet? C. What effect does the mylar sheet have on the electric field between the plates? Explain how you can reconcile thiswith the increase of charge on the plates, which acts to increase the electric field. Answer α) δQ = (2.1 - 2.5 - 107 - 12 = 6.3 - 10-6 С b) Q₁=Q(1-1) k = (9.3-10-6). (1-1) 3.1 = 6.3.10-6 C c) The addition of themylar does not affect the electric field since the induced charge cancels the additional charge drawn to the plates.