PH04/PH4S MECHANICS SEMESTER I EXAMINATION 06-07 SOLUTION MULTIPLE-CHOICE QUESTIONS. (B) For freely falling bodies, the equation v = gh holds. v is proportional to h, therefore v v = h h = h h =.. (B).5i + 0.3j From the scalar product of vectors A and B, we get A B =.00x b + 3.00y B = 3.00 x b = 3.00 3.00y B..00 Because the angle between A and B is 45.0, the scalar product can also be written as A B cos 45.0 = 3.00 3.0 x b + y B = 3.00 Substituting x b from the previous equation, y B = 0.3 or y B =.5 Choose y B = 0.3 and x b =.5 because the x-component of vector B is positive. 3. (B) 390N The object does not move perpendicularly to the ground, therefore F y = 0 F sin 30 mg + N = 0 N = 390N 4. (C) 5N Forces on the upper block: F T mg =.3 () Forces on the lower block: T T mg =.3 () () + () and inserting T = m rope g: Back substitution to (): F = 4.6 + (m rope + m)g T = (m rope + m)g +.3 = 5N 5. (A) 4. x 0 3 N
In the plane s frame of reference, the pilot is experiencing a fictitious centrifugal force and not moving perpendicular to the floor, F y = 0 N = mg + F centrifugal = 4. 0 3 N 6. (B) 00 m/s All the kinetic energy after the collision between the bullet and the block is converted into the spring s potential energy mv = kx v =.ms From conservation of momentum in the event of collision we know, m bullet v bullet = (m block + m bullet )v v bullet = 00m/s 7. (D).67R The race car leaves the track when there is no normal force from the track onto the car. N = 0 mv R = mg sin θ v = gr sin θ Since there is no external work on the system, conservation of energy yields KE + PE = KE + PE 0 + mg(r) = mgr sin θ + mgr( + sin θ) sin θ = 3 The height of the car is h = R( + sin θ) =.67R 8. (C) the momentum change of the lighter fragment is exactly the same as the momentum change of the heavier fragment. We can apply the conservation of momentum here, inserting zero to the initial momentum as the two fragments were not moving relative to each other. 0 = p + p p = p p = p 9. (C) 3.67 m/s The work done on the object W = U = U(.00) U(5.00) = 56 5 The kinetic energy of the object when it is located at x = 5.00m. KE = KE + W mv = mv + 56 5 v = 3.67m/s
0. (C).7cm The initial condition when the block is held in place against the spring F = kx k = 74 x () Apply conservation of energy at the initial condition and at the time the block is released kx = mv 74x =.37(.) x =.7 0 m. (A) 0.7m/s Tangential acceleration a t = αr = 0.m/s Radial acceleration a r = ω r = (αθ)r = 0.7m/s Total linear acceleration a = a r + a t = 0.7m/s. (C) 3gl Use conservation of energy, i.e. all the gravitational potential energy is converted into rotational kinetic energy Iω = Mg h Since the rod is rotating around one of its end, we know I = 3 ML and ω = v tip L h is the center of mass change of height, which is equal to L for thin enough rod (shown in the figure on the right) We then obtain v tip = 3gL 3. (C).76 x 0-3 s Simply apply the conservation of angular momentum law Iω = I ω Assume the shape remains the same (spherical), therefore I can be expressed as a function of its radius, i.e. I(r) = kmr. kmr ω = kmr ω ω = ( r r ) ω The period is T = π ω = T (r r ) =.76 0 3 s
4. (A) Axis Moment of inertia is proportional to radius (it is more difficult to rotate a body that is farther away from the axis) 5. (D) Search for gyroscopic stabilization and tail rotor for more information. 6. (B) 3g Surface gravity is proportional to the body s mass and inversely proportional to its radius squared g M and g R g = 3 M M ( R ) g = 3g 3 R 7. (A).99 x 0 4 kg We have two equations with two unknowns (R and M). At the surface, G Mm = 389N () R At.86 x 0 4 km above the planet s surface, Mm G (R +.86 0 4 = 4.3N () km) Dividing () by () (R +.86 0 4 km) R = 389 R = 600km 4.3 Inserting G = 6.67 x 0 - N m kg -, m = 75.0 kg, and R = 6. x 0 6 m, we get 389N R M = =.99 0 4 kg Gm 8. (A) (a)7. x 0 5 kg, (b).9 x 0 30 kg The surface gravity of a planet is related to its mass(m) and radius(r) but not its rotation period (.3 hours is not needed) g = Gm m = 7. 05 kg r The dynamics of the planet s revolution is governed by the circular motion equation F centripetal = F gravity mω R = GMm R M is the mass of the star, R is the orbital radius of the planet, ω = π where T = 40 earth days. T M = 4π R 3 G T =.9 030 kg 9. (A).8 x 0 4 kg
From the orbital speed equation v orbit = GM, we can represent the radius of the planet in R+h terms of the known variables (G, v orbit, h) and the mass. R = GM h v orbit We can then substitute this R into the equation for escape velocity ( v escape v orbital v e = GM R h ) M = vescape G 0. (E) 6.4 x 0 8 m/s The magnitude of acceleration due to gravity from each mass a = GM R = 3.7 0 8 m/s Since the masses lie on the corners of an equilateral triangle, we know the angle between the two acceleration vectors is 60. Therefore, the total acceleration is a total = a ( + cos 60 ) = 6.4 0 8 m/s. #a Since the plane is only moving horizontally, the kinematics equation for the vertical position of the bomb is When the bomb hits the ground, y = h gt y = 0 t = h g For the x(horizontal) coordinate, the bomb has the same constant velocity as the plane. Simply substitute the time t into the kinematics equation for movement constant velocity. x = vt = v h g #b (a) Viewed in a frame of reference that is moving with the cart, the kg box will experience a fictitious force to the opposite direction from the acceleration. The magnitude of this fictitious force is F = ma where m is the mass of the box and a is the acceleration of the cart. The box must remain stationary in the moving frame of reference for it to not fall (in the non-moving frame of reference: moving forward together with the cart and not moving vertically).
Newton s equation for the box in this frame of reference is then: F x = 0 N = F N = ma We know that normal force affects the maximum magnitude of static friction, i.e. f s,max = μ s N. From these two relations, we can think of finding the minimum acceleration as finding the minimum value of maximum static friction, i.e. when the maximum static friction is just enough to counter the box s weight. (b) 0N. f s,max = mg μ s ma = mg a = g μ s = 6m/s (c) 0N. Frictional force always resists the tendency of movement, but never creates movement. Here, static frictional force only nullify gravity and will not exceed the weight. (d) a g μ s, f s,max = μ s m(a) μ s m( g μ s ) = mg The box will not fall #a On the radial direction(r) the string is always tense and will not increase or decrease in length: a r = 0 On the tangential direction(t): F t = ma t a t = g sin θ (i) (ii) (iii) The acceleration of ball A after it is released a A = a t = 9.8 sin 30 = 4.9m/s The acceleration of ball A just before colliding with ball B a A = a t = 9.8 sin 0 = 0m/s Ball A stops, and ball B moves with the same velocity as ball A upon the impact. Ball B then instantaneously collides with C, and C with D. Finally, D moves upward as high as A initially. This happens because the collision is elastic, hence kinetic energy is conserved. One example is perfectly non-elastic collision where all the balls stick and move together. Linear momentum is conserved if the collision is non-elastic, but energy is not. m A v A = (m A + m B + m C + m D )v m Av A (m A + m B + m C + m D )v
#b Because the snow is frictionless, there is no energy loss along the ramp. From conservation of energy: v = g h = 9.8 (5 3.0) = m/s The rest of the steps required to find the touchdown point is simply the kinematics of parabolic motion. We know that she launches off from the ramp at an angle of 30 from the ground, so The uptime is t up = 9.8 =.s v x = cos 30 = 8m/s and v y = sin 30 = m/s The maximum height and the downtime are h max = 3 +. 9.8. = 9.m and t down = h max g = (9.) 9.8 =.4s The distance to her touchdown point is then x = v x (t up + t down ) = 8(. +.4) = 45m #3a (i) (ii) There is slipping between the spool and the incline. The string is pulling the spool to rotate backwards while the spool itself is moving forward down due to gravity. Translationally, the spool is not moving perpendicularly to the incline, so a = 0. Parallel to the incline, the ball is moving because of the forces F = ma ma = mg sin θ T a = g sin θ T m. (iii) Substituting α = a r Rotationally the spool is only rotating because of the tension: Iα = Tr α = Tr I into the equations from (ii), we get a = g sin θ Ia mr a ( + I mr) = g sin θ g sin θ a = + I mr #3b The static friction resists the tendency of movement which is downward. Therefore, the direction of the static frictional force is upward parallel to the incline. Since the spool is stationary, we know
F = 0 T = mg sin θ f () τ = 0 fr Tr = 0 T = fr r () Equating T from () and () mg sin θ f = fr r f() = mgr sin θ f = r mg sin θ To find the minimum coefficient of static friction we take the maximum static frictional force to be equal with the frictional force we have found, i.e. f s,max = μ s N = r mg sin θ r mg sin θ The normal force can be expressed in terms of m, g, and θ by considering the forces along the perpendicular direction F = 0 N = mg cos θ μ s mg cos θ = μ s = r mg sin θ tan θ