Tiêu đề 4.MOTION IN A PLANE - Explanations.pdf ✅

1 (c) tan θ = v 2 rg ∴ θ = tan−1 ( v 2 rg) = tan−1 ( 10 × 10 10 × 10) ∴ θ = tan−1(1) = 45° 2 (b) From, tan θ = v 2 rg r = v 2 g tan θ = 10 × 10 10 × tan 45° = 10 m 3 (c) L = Iω. In U. C. M.ω = constant ∴ L = constant. 5 (a) ω = v r = 100 100 = 1 rad/s 6 (b) Here, A1 = A2 = 1 and A1 2 + A2 2 + 2A1A2cosθ = (√3) 2 = 3 or 1+1+2× 1 × 1 ×cosθ = 3 or cosθ = 1 2 Now, (A⃗ 1 − A⃗ 2) ∙ (2A⃗ 1 + A⃗ 2) = 2A1 2 − A2 2 − A1A2 cosθ = 2 × 1 2 − 1 2 − 1 × 1 × 1 2 = 1 2 7 (a) A⃗ × B⃗ = (4î+ 6ĵ) × (2î+ 3ĵ) = 12(î× ĵ) + 12(ĵ× î) = 12(î× ĵ) − 12(î× ĵ) = 0 Again, A⃗ ∙ B⃗ = (4î+ 6ĵ) ∙ (2ĵ+ 3î) = 8 + 18 = 26 Again, |A⃗ | |B⃗ | = √16+36 √4+9 ≠ 1 2 Again , B⃗ = 1 2 A⃗ 8 (c) Difference in K.E. = Difference in P.E. = 2mgr 9 (a) We know that tan θ = v 2 Rg and tan θ = h b Hence h b = v 2 Rg ⇒ h = v 2b Rg 10 (b) Since F⃗ = 4î− 3ĵis lying in X − Y plane, hence the vector perpendicular to F⃗ must be lying perpendicular toX − Y plane ie, along Z-axis. 11 (d) Angular momentum is an axial vector. It is directed always in a fix direction (perpendicular to the plane of rotation either outward or inward), if the sense of rotation remain same 12 (a) Here, v⃗⃗⃗1 =30km h −1 due north = OA⃗⃗⃗⃗⃗ v⃗ 2 = 40kmh −1 due east = OB⃗⃗⃗⃗⃗ Change in velocity in 20 s ∆v⃗ = v⃗ 2 − v⃗⃗⃗1 = v⃗ 2 + (⃗−⃗⃗⃗⃗v⃗⃗1 ) = OB⃗⃗⃗⃗⃗ + OC⃗⃗⃗⃗⃗ = OD⃗⃗⃗⃗⃗ |∆v⃗ | = √v2 2 + v1 2 = √402 + 302 = 50 kmh −1 Acceleration , a⃗ = |∆ v⃗ | ∆t = 50 20 = 2.5 kmh −2 Tan β = v1 v2 = 30 40 = 0.75 = tan37° ∴ β = 37° north of east   g v 2 /r
14 (d) We know that R = u 2 sin 2 θ g = 10 × 10 × sin60° 10 = 10 × √3 2 = 5 × 1.732 = 8.66 m 15 (c) Angular momentum L = r × p = r × m × v v = L mr ... . (i) Now, as centripetal force, Fc = mv2 r ... . (ii) Substituting the value of v from Eq. (i) in Eq. (ii), we get Fc = m r [ L mr] 2 = L 2 mr3 16 (a) Maximum height, H = u 2 sin2θ 2g Range, R = u 2 sin2θ g Given, H = R 2 ∴ u 2 sin2θ 2g = u 22sinθcosθ 2g or sinθ = 2cosθ or tanθ = 2 or θ = tan−1 (2) 17 (a) Max. tension that string can bear = 3.7 kgwt = 37 N Tension at lowest point of vertical loop = mg + mω 2 r = 0.5 × 10 + 0.5 × ω 2 × 4 = 5 + 2ω 2 ∴ 37 = 5 + 2ω 2 ⇒ ω = 4 rad/s 18 (a) v = K(yî+ xĵ) vx = Ky dx dt = Ky Similarly, dy dt = Kx Hence dy dx = x y ⇒ y dy = x dx , by integrating y 2 = x 2 + c 19 (d) R = u 2 sin 2θ g ∴ R ∝ u 2 . If initial velocity be doubled then range will become four times 20 (d) v = kyi̇ ̂ + kx j̇ ̂ dx dt = ky, dy dt = kx dy dx = dy dt × dt dx = kx ky ydy = xdx y 2 = x 2 +c 21 (d) Tension in the string T = mω 2 r = 4π 2n 2mr ∴ T ∝ n 2 ⇒ n2 n1 = √ T2 T1 ⇒ n2 = 5√ 2T T = 7 rpm 22 (b) h = (u sin θ)t − 1 2 gt 2 d = (u cos θ)t or t = d u cos θ h = u sin θ . d u cos θ − 1 2 g. d 2 u 2 cos2 θ u = d cos θ √ g 2(d tan θ − h) 23 (a) Using v 2 − u 2 = 2as, we get s = v 2 2g Now, v 2 sin 2θ g = v 2 2g or sin 2θ = 1 2 or sin 2θ = sin 30° or θ = 15°
The other possible angle of projection is (90° − 15°), ie, 75° 24 (b) R = 2 × 30 × 30 sin 30° cos 60° 10 cos2 30° = 180 × 1 2 × 1 2 × 2 × 2 3 m = 60 m 25 (c) For a particle in uniform circular motion a = v 2 R = towards center of circle a = v 2 R (− cos θ i̇ ̂ − sin θ j̇ ̂) or a = v 2 R cos θ i̇ ̂ − v 2 R sinθ j̇ ̂ 26 (c) α < β if B⃗ < A⃗ or B < A 27 (d) In a circular motion a = v 2 r ⇒ a2 a1 = ( v2 v1 ) 2 = ( 2v1 v1 ) 2 = 4 28 (c) Due to constant velocity along horizontal and vertical downward force of gravity stone will hit the ground following parabolic path 29 (a) At the highest point, velocity is horizontal 30 (a) Balancing the force, we get Mg − N = M v 2 R For weightlessness, N = 0 ∴ Mv2 R = Mg or v = √Rg Putting the values, R = 20 m, g = 10.0 ms−2 So, v = √20 × 10.0 = 14.14 ms−1 Thus, the speed of the car at the top of the hill is between 14 ms−1 and 10 ms−1 31 (c) When a stone tied at the end of string is rotated in a circle, the velocity of the stone at an instant acts tangentially outwards the circle. When the string is released, the stone files off tangentially outwards ie, in the direction of velocity 32 (c) Kinetic energy = potential energy 1 2 m(kve ) 2 = mgh 1 + h R ⟹ 1 2 mk 22gR = mgh 1 + h R ⟹ h = Rk 2 1 − k 2 Height of projectile from the earth’s surface = h Height from centre r = R + h = R + Rk 2 1 − k 2 By solving r = R 1 − k 2 33 (a) Retarding force F = ma = μR = μ mg a = μg
Now, from equation of motion, v 2 = u 2 − 2as ∴ 0 = u 2 − 2as ∴ s = u 2 2a = u 2 2μg = v0 2 2μg 34 (c) If a particle is projected with velocity u at an angle θ with the horizontal, the velocity of the particle at the highest point is v = u cos θ = 200 cos 60° = 100 ms −1 If m is the mass of the particle, then its initial momentum at highest point in the horizontal direction = mv = m × 100. It means at the highest point, initially the particle has no momentum vertically upwards or downwards. Therefore, after explosion, the final momentum of the particles going upwards and downwards must be zero. Hence, the final momentum after explosion is the momentum of the third particle, in the horizontal direction. If the third particle moves with velocity v ′ , then its momentum = mv′ 3 , According to law of conservation of linear momentum, We have mv′ 3 = m × 100 or v ′ = 300 ms −1 35 (b) Net acceleration in nonuniform circular motion, a = √at 2 + ac 2 = √(2) 2 + ( 900 500) 2 = 2.7 m/s 2 at = tangential acceleration ac = centripetal acceleration = v 2 r 36 (a) |P⃗ | = A + B ⇒ |P⃗ | 2 = (A + B) 2 |A⃗ + B⃗ | 2 = (A + B) 2 or A 2 + B 2 + 2AB cosθ = A 2 + B 2 + 2AB or cosθ = 1 or θ = 0° 37 (d) Maximum tension in the string is Tmax = mg + mvB 2 l = mg + 2mgl l (1 − cos θ0) = mg + 2mgl l . 2 sin2 θ0 2 ∵ (1 − cos θ0 = 2 sin2 θ0 2 ) [Since θ0 is small] = mg(1 + θ0 2 ) 38 (d) Angular acceleration = d 2θ dt 2 = 2θ2 39 (a) Extension is first case ∆l1 = 2a − a = a From Hooke’s law F = k∆l1 mrω 2 = ka m(2a)ω 2 = ka 2ma ( 2π T ) 2 = ka In second case, m(3a) ( 2π T ′ ) 2 = k(3a − 2a) On dividing Eq. (ii) by Eq. (i) T ′ = √ 3 2 T 40 (b) New kinetic energy E ′ = E = cos2θ = Ecos2(45°) E 2 41 (a) F = mv 2 r or v = √ Fr m