Title 5.LAWS OF MOTION - Explanations.pdf ✅

1 (a) In case of upward motion F = m(g + a) = 60(9.8 + 4.9) = 60(14.7) = 882 kg 2 (a) Change in momentum = Impulse ⇒ ∆p = F × ∆t ⇒ ∆p F = 125 250 = 0.5 sec 4 (d) Here : Mass of ship m = 2 × 107 kg, Force F = 25 × 105N Displacement s = 25 m According to the Newton’s second law of motion F = ma ⇒ a = F m = 25 × 105 2 × 107 = 12.5 × 10−2 m/s 2 The relation for final velocity is v 2 = u 2 + 2as ⇒ v 2 = 0 + 2 × (12.5 × 10−2) × 25 ⇒ v = √6.25 = 2.5 m/s 5 (d) If rope of lift breaks suddenly, acceleration becomes equal to g so that tension, T = m(g − g) = 0 6 (b) Now, atB In equilibrium, T cos θ = mg ⟹ cos θ = 150 × 9.8 2940 ⟹ cos θ = 0.5 ⟹ θ = 60° 7 (b) When two blocks performs simple harmonic motion together then at the extreme position (at amplitude = A) Restoring force F = KA + 2ma ⇒ a = KA 2m There will be no relative motion between P and Q if pseudo force on block P is less than or just equal to limiting friction between P and Q i. e. m = ( KA 2m ) = Limiting friction ∴ Maximum friction = KA 2 8 (a) The arrangement in shown in figure Now, draw the free body diagram of the spring balances and block. For equilibrium of block, T1 = Mg Where, T1 = Reading of S2 For equilibrium of S2, T2 = T1 Where, T2 = Reading of S1 For equilibrium of S1, T2 = T3 Hence, T1, = T2 = Mg So, both scales read M kg. 9 (c) Given that, v1 = 10 ms−1 ; m1 = 10 kg v2 = 0; m2 = 9 kg v3 = v; m3 = 1 kg According to conservation of momentum m1v1 = m2v2 + m3v3 10 × 10 = 9 × 0 + 1 × v ⟹ v = 100 ms−1 10 (a) For equilibrium of forces, the resultant of two (smaller) forces should be equal and opposite to third one 11 (c) At 11th second lift is moving upward with acceleration a = 0 − 3.6 2 = −1.8 m/s 2 Tension in rope, T = m(g − a) = 1500(9.8 − 1.8) = 12000 N 12 (c) T = m(g + a) = 1000(9.8 + 1) = 10800 N 13 (c) Force applied by engine= 6 m When two cars are pulled, (m + m)a = 6 m
or 2ma = 6m or a = 3 ms −2 14 (b) From the relation, acceleration a = F m1 + m2 + m3 ⇒ a = 40 10 + 6 + 4 = 2ms −2 ∴ 40 − T = 10 × 2 T = 20 N 15 (c) If T is tension in each part of the string holding mass √2m, then in equilibrium, T cos θ + T cos θ = √2 mg 2 T cos θ = √2 mg But T = mg; ∴ 2mg cos θ = √2mg cos θ = 1 √2 θ = 45° 16 (b) As both the balls are of same size, force of buoyancy on each is same. Therefore, in equilibrium, F + F = m1g + m2g or F = (m1 + m2 ) g 2 Considering the equilibrium of lower ball, T + F = m1g T = m1g − F T = m1g − (m1 + m2 ) g 2 T = (m1 − m2 ) g 2 17 (a) Mass of the rocket m = 1000 kg ∆m ∆t = 4 kgs−1 , v = 3000 ms−1 Thrust on the rocket F = −v ∆m ∆t = −3000 × 4 = −12000 N (Negative sign indicates that thrust applied in a direction opposite to the direction of escaping gas) 18 (d) Rate of flow of water V t = 10 cm3 sec = 10 × 10−6 m3 sec Density of water ρ = 103kg m3 Cross-sectional area of pipe A = π(0.5 × 10−3) 2 Force = m dv dt = mv t = Vρv t = ρV t × V At = ( V t ) 2 ρ A (∵ v = V At) By substituting the value in the above formula we get F = 0.127 N 19 (a) Let R be the reaction of base of lift and g the acceleration due to gravity, acting downwards. Their resultant provides the net acceleration to the lift. Therefore, R − mg = ma ⟹ R = m(g + a) Also, R = mg ′ ⟹ g ′ = g + a Therefore, net acceleration increases and hence, reading indicated by spring balance will increase. 20 (b) We know s = u 2 2μ g ∴ μ = u 2 2gs = (6) 2 2×10×9 = 0.2 21 (c) F = dp dt = v ( dM dt ) = αv 2 ∴ a = F M = αv 2 M 22 (d) Acceleration of the system a = F (m + m 3 ) ∴ Tension in the middle of the rope T = (m + m 6 ) a or T = 7m 6 ∙ 3F 4m = 7 8 F 23 (a) According to Newton’s second law : Force = rate of change of linear momentum 24 (d) As the springs are light in weight, therefore, tension in both springs will be same. So, both springs will show same reading 8 kg. 25 (d) Horizontal velocity of ball and person are same as both will cover equal horizontal distance in a given interval of time and after following the parabolic path the ball falls exactly in the hand which threw it up 26 (c) Mass of the packet M = w/g For packet F − w = Ma
or F − w = ( w g ) × 2g or F = 3w 27 (c) Newton’s first law of motion defines the inertia of body. It states that every body has a tendency to remain in its state (either rest or motion) due to its inertia 28 (b) aA = g/2 aB = g 29 (a) μ = tan θ (1 − 1 n 2 ) = tan θ (1 − 1 2 2 ) = 3 4 tan θ 30 (c) Direction of second force should be at 180° 32 (d) Force acting on plate, F = dp dt = v ( dm dt ) Mass of water reaching the plate per sec = dm dt = Avρ = A(v1 + v2 )ρ = V v2 (v1 + v2)ρ ( v = v1 + v2 = velocity of water coming out of jet w. r.t. plate) [A = Area of cross section of jet = V v2 ] ∴ F = dm dt v = V v2 (v1 + v2 )ρ × (v1 + v2 ) = ρ [ V v2 ] (v1 + v2 ) 2 34 (a) Retardation of train = 36 kmh −1 5 s = 35 × 5 18 ms −1 5s = 2 ms −2 It acts in the backward direction Fictitious force on suitcase = 2m N, Where m is the mass of suitcase It acts in the forward direction Due to this force, the suitcase has a tendency to slide forward. If suitcase is not to slide, then 2m = force f of friction or 2m = μmg or μ = 2 g = 2 9.8 = 20 98 = 10 49 35 (d) F1 = mg(sin θ + μ cos θ) F2 = mg(sin θ + μ cos θ) F1 F2 = sin θ + μ cos θ sin θ − μ cos θ = tan θ + μ tan θ − μ = 2 μ + μ 2 μ − μ = 3 36 (a) F = W μ ∴ W = μF = 0.2 × 10 = 2N 37 (c) l = l0√1 − v 2 c 2 = 1√1 − ( 2.7 × 108 3 × 108 ) 2 ⇒ l = 0.44 m 38 (d) Time taken by the bullet and ball to strike the ground is t = √ 2h g = √ 2 × 5 10 = 1 s Let v1 and v2 are the velocities of ball and bullet after collision. Then applying x = vt We have, 20 = v1 × 1 or v1 = 20 m/s 100 = v2 × 1 or v2 = 100 m/s Now, from conservation of linear momentum before and after collision we have, 0.01v = (0.2 × 20) + (0.01 × 100) On solving, we get v = 500 m/s 39 (a) When friction absent a1 = g sinθ ∴ s1 = 1 2 a2t2 2 ... . (i) When friction in present a2 = g sin θ − μkg cos θ ∴ s2 = 1 2 a2t2 2 ... . (ii) From Eqs. (i) and (ii), we have 1 2 a1t1 2 = 1 2 a2t2 2 or a1t1 2 = a2(nt1) 2 (∵ t2 = nt2) or a1 = n 2a2 or a2 a1 = g sin θ − μkg cos θ g sin θ = 1 n 2 or g sin45° − μkg cos 45° g sin 45° = 1 n 2 or 1 − μk = 1 n 2 or μk = 1 − 1 n 2 40 (c) vG = mBvB mG = 0.2 × 5 1 = 1 m/s
41 (b) If v is common velocity of the block and movable wedge, then applying the principle o conservation of linear momentum we get, mu + 0 = (m + nm)v v = mu m(1 + n) = u 1 + n This infact, can be taken as velocity of centre of mass of the block and wedge ie, vCM = v = u 1 + n Applying the principle of conservation of energy 1 2 mu 2 = mgh + 1 2 m(1 + n) u 2 (1 + n) 2 or u 2 = 2gh + u 2 1+n u = √2gh (1 + 1 n ) 42 (b) Acceleration produced in jet = Change in velocity Time a = (103 − 0) 10 = 100 m/s 2 ∴ Mass = Force Acceleration = 105 102 = 103 kg 43 (c) If 1 N and 2 N act in the same direction and 3 N acts in opposite direction, equilibrium is possible 44 (c) Force = v dm dt Force = v M [∴ Force = M v newton] 45 (c) For given condition we can apply direct formula l1 = ( μ μ + 1 ) l 46 (c) If monkey move downward with acceleration a then its apparent weight decreases. In that condition Tension in string = m(g − a) This should not be exceed over breaking strength of the rope i. e. 360 ≥ m(g − a) ⇒ 360 ≥ 60(10 − a) ⇒ a ≥ 4 m/s 2 47 (c) a = m2 − m1 m1 + m2 g = 10 − 5 10 + 5 g = g 3 48 (d) Force equation for ‘M’ Ma = T ...(i) Force equation for, m mg − T = ma ...(ii) On solving (i) and (ii) T = ( Mm M + m ) g 49 (a) Tension, thrust, air resistance, weight are all common forces in mechanics whereas impulse is not a force Impulse = Force × Time duration 50 (a) f = μmg = 0.8 × 4 × 10 = 32 N Applied force F < f therefore, answer will be (a) 51 (b) Resolve momentum 6.5 m along x and y axes and equate ∴ 6.5 m cos θ = 5 × 1 and 6.5 m sinθ = 6 × 2 ⇒ (6.5 m) 2 = (5) 2 + (12) 2 ⇒ 6.5m = 13 ⇒ m = 2 kg ∴ Total mass = 1 + 2 + 2 = 5kg 52 (a) Since, P = (M + m)a Now as in free body diagram of block, ma cos α = mg sin α ∴ a = g sin α cosα = g tan α or P = (M + m)g tan α 53 (c) For given condition s ∝ 1 m2 ∴ s2 s1 = ( m1 m2 ) 2 = ( 200 300) 2 ⇒ s2 = s1 × 4 9 = 36 × 4 9 = 16 m 54 (d) K = F x and increment in length is proportional the original length i. e. x ∝ l ∴ K ∝ 1 l It means graph between K and l should be hyperbolic in nature 55 (c) The frictional force F is given by F = μk mg Given, μk = 0.4, m = 2 kg, g = 10 ms−2