RI Y5 Remedial Dynamics.1 Soln

Raffles Institution Physics Department Y5 H2 Physics Term 3 Remedial 8 Chapter 3: Dynamics (Newton’s Laws of Motion) Solutions 1         Apply Newton's 2nd law to the horizontal motion of the system of three blocks: total mass of system 2 3 4 2 18 N Next, apply Newton's 2nd law to the horizontal motion of the 4 kg block: 44 2 18 Fa FT a T         81 0 N ANS: C 2 Using the coordinate system shown, the resultant force along the direction is zero: sin cos 0 the resultant force along the direction is: cos sin y x yF N F m g xF F m g       ANS: A 3 Consider all the masses together, and let N be the normal contact force from floor of lift on the 5M: N – 9Mg = (9M) a N = 9Ma + 9Mg Consider forces acting on the 5M mass and let R be the force M exerts on 5M : N – 5Mg – R = (5M) a R = N – 5Mg – 5Ma = (9Ma + 9Mg) – 5Mg – 5Ma = 4 Ma + 4 Mg ANS: C F mg N θ θ  θ

Raffles Institution Physics Department Y5 H2 Physics Term 3 Remedial 9 4 (a) (b) Rock : (10)(9.81) – T = 10 a Box and ball: T = (35) a Adding the two equations: (10)(9.81) = (10+35) a a = 2.18 m s–2 Ball : T sin  = ma T cos  = mg Dividing the two equations : tan  = a / g = 2.18 / 9.81  = 12.5  (c) The maximum angle is 45° when the acceleration of the system of rock, box and ball is at its maximum value of 9.81 m s2. This value of acceleration can occur if the mass of the rock is very large compared to the mass of the box and ball. (refer to working in (b) : mrock g = (mrock + mbox + mball) a t a n θ = g g =1 θ = 45° 5 (a) (i) Consider forces along the slope: F1 cos 30° = (40)(9.81) sin 30° F1 = 227 N (ii) consider forces perpendicular to slope: N = F1 sin 30° + (40)(9.81) cos 30° = 453 N (b) F = ma (40)(9.81) sin 30° = 40 a a = 4.91 m s–2 (c) If friction is not negligible: F1cos300 + friction = mgsin300. Thus to maintain equilibrium, the value of F1 will be smaller.