LOM-08-ANSWERSHEET

1. (a) 2. (b) 3. (c) 4. (c) 5. (a) 6. (a) 7. (c) 8. (b) 9. (b) 10. (c) 11. (b) 12. (c) 13. (b) 14. (a) 15. (b) 16. (c) 1. (b) 2. (d) 3. (a) 4. (a) 5. (c) 6. (d) 7. (d) 8. (a) 9. (a) 10. ( d) 1. (a) 70 kg (b) 35 kg (c) 105 kg (d) zero. 2. In all the three intervals, acceleration and, therefore forces are zero. 3. 0.05 x 12 = 0.6 kg ms-1 (for each ball) 4. Net force = 65 kg x 1 ms-2 = 65 N amax = sg  2ms2 5. (a) 750 N (b) 250 N Mode (b) should be adopted. 6. (a) (c) T = 640 N T = 400 N (b) (d) T = 240 N T = 0. The rope will break in case (a) 7. 20 m 8.   60º 9. (a) 7500 N downward (b) 32500 N downward. (c) 32500 N upward 10. (a) accelerated with acceleration 0.5 m/s2 (b) at rest 11. 11. (K1  K2 )x m 12. g/3 T  2 mg  4 g . 3 3 13.   tan 1 v2  rg .   14. T  mg tanθ , T  mg , T '  mg cosθ 15. ab ob a1  m2 cosθ ob a2 m1 SUBJECTIVE LEVEL - II Mg 1. 4h d2  4h2 2. 4 N  Mg  B  3. 2   g  4. (a) 3/7 g (b) motion of both the blocks is independent a  F1  m1g  12m / s2 . m1 5. (a) F  (k1  k2 )m1m2 g cosα m1  m2 (b) tanα min  k1m1  k2 m2 m1  m2 1 1 1  6.   tan      7. (a) mg 2 cosα v  2a sin2 α (b) m2 g 3 cosα s  6a2 sin3 α 8. (a) f  3 μMg ; A 4 fB  0; T  0 (b) f  μMg ; T  μMg ; f A 2 B  μMg 2 (M  m)g 9. 1   10. w2  g(η  sinα  k cosα ) /(η  1)  0.05 g SUBJECTIVE LEVEL - III 1. a  a  3 gsin  a  gsin  T  Mgsin  . A C 4 B 8 (M  m) tan  2. 1  tan  3. 150 N, 30 m/s2. 4. t  5. aA  6. g 1 η cot2 α . ; aB  g tanα η cotα 7. tan   Tmin mg(sin   cos ) . f  2l Mm (M  m)t 2 9. when t  t0 , the accelerations w1  w2  at /(m1  m2 ); when t  t0 w1  kgm2 / m1 , w2  (at  km2 g) / m2 Here, t0  kgm2 (m1  m2 ) / am. 10. τ  PROBLEMS ANSWERS 1. → ˆ ˆ PPM  mv 2 sin t i  m(v2 cos t  v1 ) j 2. min  9.89 rad / sec. 3. (b) (i) f1  30N f2  15N (ii) for for for m1, 30  T  20a m2 , T  15  5a m, F  30  50a . (iii) F  60N, T  18N, a  3 / 5 m / s2 4. (i) f = 36 N (ii) min  11.67 rad / s (iii) r1  0.1m, r2  0.2 m .

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