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1998 Mechanics C1 Solution

1998 Mechanics C1 Solution - Distribution(a of points i 1...

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Unformatted text preview: Distribution (a) of points i. 1 point For correct answer 1 point _ As 0.30m—0.10m I) =— =——-—-———=1.00m/s At 0305—0103 ii. 1 point For correct answer 1 point B=£_ 0.99m—0.87m 2060mm At ‘ 1.10s-0.90s iii. lpoint For correct answer 1 point U=SA_s_:1.18m—1.14m =O'20m/s At 1.908—1.7OS (b) 3 points v(m/s) L50 _fl_-fi__fi__+__r_q--_P-1__+__+__h-_r_q__*__r__r__r_fi__fi__1 __I___|.___I _ _ _ . __.'___I__ __l__.I..._l__._L.__I___l__L__i___I___|_-J__.I I I I I I I I I I I I I I I —4—-4-~4——+——F——H——h—4-—+--+——F——F—4——4——+——+-—F—A——4——4 I I I I I I I I I I I I I I —j__j_'1_“T—"r——__—r'1_‘T-~T__F~_F_j__7_—r'—T_“F_7__7__1 I .I__L_._J._..I___L.__I___l_..L_—L——|_——I_—_I__J I I I I I I I I | I I I I __T——T‘-I'__I'_'1“"I"T'-‘T__I'_"I""I-‘1 _ _ _ _ _ _ . _ _ _ _ _ _ _ _ -_ L_-L__L__L_J__J__L__L_-L_J__J__J | | | | I I I I I I 050 '1“fi"1"T"r"F"F'1'-T"T‘ r"r'fi"1--r—‘T"F—1"1-‘1 _4__J-_J__I__L_4__4__J-_i-_L__ _L_J__J-_L__L_ _J__J__J I I I I I I I I I I I I I I I I I I I -fl--fi--1--r--r--r--r-1--T--r--r- ‘fi_—1-—f' -r-fi*-fi--1 _J__J__J__l__L__L_-L-J__l__L__L__L_ I I I I I I I I I I I I ——h-4——4——+-—F——F——F—4--+——+--F-- 0 0.50 1.00 1.50 ' 2.00 For line I horizontal at v = 1.00 m/s or at answer obtained for (a)i. 1 point For line 11 with monotonic, negative slope between points P1 and P2, P1 at (0.70 - 0.80, 1.00) o_r (0.70 - 0.80, answer to (a)i.), and P; at (1.20 - 1.30, 0.20) o_r (1.20 - 1.30, answer to (a)iii.) 1 point (This line may be straight, which is consistent with the data, or slightly curved as shown, which is consistent with the behavior of springs) For line 111 horizontal at v = 0.20 m/s g at answer obtained for (a)iii. 1 point Distribution of points Mech, 1 (continued) (0) i. 3 points For any statement of conservation of momentum or energy 1 point For proper conservation of momentum or energy equation 1 point Method 1: Conservation of momentum mA vAi = mA vAf + mB “B (0.90 kg)(1.00 m/s) = (0.90 kg)(0.20 m/s) + (0.60 kg)vB Method 2: Recognize from part (d) that energy is also conserved. 1 2 1 2 1 2 EmAvAi =EmAvAf +EmBUE %(0.90 ng1.00 m/s)2 = i (0.90 ng020 [my +%(0.60 kg)v§ For correct answer 1 point 08 = 1.2 m/s ii. 3 points v(m/s) | I I | I I I I | I l | I I l I I I I I IIZIIIZIIILIIZIZJIIZZIIIIIii-"IIJIILflCIEIIII] I I I I I I I I I I I I ZI "'I"'"I_"'I"T':'I‘"l"—I"'I"T"T"F"F‘ 0.50 t (s) 0 0.50 8 1.00 1.50 2.00 For line I horizontal at v = O 1 point For line II with monotonic, positive slope between points P1 and P2, P1 at (0.70 - 0.80, 0), and P2 at (1.20 - 1.30, 1.20) m (1.20 - 1.30, answer to (c)i.) 1 point (This line may be straight, which is consistent with the data, or slightly curved as shown, which is consistent with the behavior of springs) For line III horizontal at 1) = 1.20 m/s 9; at answer obtained for (0)1. 1 point Distribution of points Mech. 1 (continued) ((1) i. 2 points For correct answer 1 point Yes, the collision is elastic. For any reasonable justification 1 point Examples: The final kinetic energy equals the initial kinetic energy. The spring force is conservative meaning the total energy stored equals the total energy released. The compressed spring stores and releases energy in equal amounts. (The justification point was not awarded if student answered “no” to the question.) ii. 1 point For any reasonable explanation 1 point Examples: The compressed spring stores maximum amount of kinetic energy. At time t = l s, there is maximum kinetic energy stored as potential energy. At time t = l s, the spring has maximum potential or elastic energy. ...
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