PHY317K Exam Compilation

A tennis ball launcher, which is sitting on a perfectly flat field, shoots a ball into the air at an angle of 30 degrees with respect to the horizontal direction. Ignore air resistance. The initial speed of the tennis ball is 50 m/s. In your calculations remember to take g = 10 m/s^2 If the x-axis and y-axis are defined so +x measures horizontal distance from the launcher in the direction the tennis ball is served, and +y measures vertical distance above the ground, approximately what are the components (vx, vy) of the ball's velocity when it is at the highest point of its trajectory?

(+43 m/s, 0 m/s) Answer is B since at the top the ball has only a horizontal component equal to its initial horizontal component v0x = v0*cos α = 50*cos 30◦ ≈ 50*0.866 m/s ≈ 43 m/s

The three-particle system shown below has masses m1 = 2 kg, m2 = 4 kg, m3 = 4 kg. The x and y coordinates in meters of the center of mass are:

(−9/5) m, 0 m Answer is B since: The center of a system is xCM = Σmixi/Σmi = (m1x1 + m2x2 + m3x3)/(m1 + m2 + m3) = (2 kg(3 m) + 4 kg(−4 m) + 4 kg(−2 m))/(2 kg + 4 kg + 4 kg) = −9/5 m Similarly, yCM = Σmiyi/Σmi = (m1y1 + m2y2 + m3y3)/(m1 + m2 + m3) = (2 kg(2 m) + 4 kg(2 m) + 4 kg(−3 m))/(2 kg + 4 kg + 4 kg) = 0 m

The displacement of an object with a mass of 3 kg oscillating on a spring with a spring constant of 27 N/m is given by x(t) = 2 · cos(ω · t + π/4). What is the velocity of the mass at t = π/12 seconds?

-6 m/s Answer is E since: ω = sqrt(k/m) = 3 rad/s The velocity is v(t) = −2ω · sin(ω · t + π/4) Plugging in 3 rad/s for ω and π/12 for t yields −6 m/s

The position of a particle is given by x(t) = A cos^2 (ωt), where A and ω are constants. What is the velocity of the particle at t = π/2ω ?

0 Answer is E since: v(t) = (d/dt)*x(t) = (d/dt)*A cos^2*(ωt) = −2Aω*cos(ωt)*sin(ωt) = v( π/2ω) = −2Aω*cos(ω · π/2ω)*sin(ω · π/2ω ) = −2Aω cos(π/2)*sin(π/2) = 0 (recall that cos(π/2) = 0)

Cart A of mass m = 1 kg on an air-track travels horizontally with a speed of 10 m/s and collides elastically with cart B of the same mass that is stationary. After the collision, velocities of both carts (vA, vB) are:

0 m/s, 10 m/s Answer is E. The collision is elastic, so kinetic energy and momentum are conserved Conservation of energy: 1 2m1v1^2 + 1 2m2v2^2 = E = const Conservation of momentum: m1v1 + m2v2 = p = const E and p are determined by v1,i and v2,i, and therefore v1,i and v2,i are one set of solutions Since m1 = m2, the two equations above still hold under exchange of v1 and v2 Thus, v1,f = v2,i, v2,f = v1,i.

A soccer ball of mass m is moving horizontally with a constant speed of 5 m/s, collides elastically with a wall, and bounces back with the same speed. If the average force exerted on the soccer ball was 200 N, and the collision lasted 0.01 s, the mass of the ball is

0.2 kg Answer is A because impulse is equal to change in momentum, i.e., Favg∆t = ∆p The incoming velocity is the same magnitude as the outgoing, but with opposite sign, meaning ∆p = m(vf − vi) = m(vf − (−vf )) = 2m|v| Thus Favg∆t = 2m|v| Solving for m gives m = Favg∆t/(2|v|) = 200 N·0.01 s/2·5 m/s = 0.2 kg.

A pendulum swings with an angular frequency of 12 rad/s. It moves with simple harmonic motion as described by θ(t) = θmax cos(ωt + φ). If its maximum angular velocity is 3.6 rad/s, what is the amplitude θmax of these oscillations?

0.3 rad Answer is B since: The equation of motion is θ(t) = θmax cos(ωt + φ) The angular velocity is the derivative, dθ(t)/dt = −Aω · sin(ωt). The maximum speed is |(dθ(t)/dt)max|= Aω. Solving for amplitude A = |(dθ(t)/dt)max|/ω = 3.6/12 = 0.3 rad

The exact amount of purely horizontal force necessary to get a wardrobe to start moving on a specific surface is 500 N. The mass of this wardrobe is 150 kg. The coefficient of static friction between the wardrobe and the surface is about

0.33 Answer is A since: The wardrobe will just start moving with a force of 500 N, so that is the maximum force of static friction. So, Frequired = fs We know that fs = µs*n = µs*mg; so Frequired = fs = µs*n = µs*mg This means that µs = Frequired/mg = 500 [N]/(150 [kg]*10 [m/s 2]) = 0.33

The ice giant Dendor has a mass of about 4 times the mass of Earth and a radius of about 5 times the radius of Earth. An object is released from rest at a height above the surface of Dendor equal to 1 radius of Dendor. The acceleration of the object is about:

0.4 m/s^2 Answer is C since: g = GMearth/Rearth^2 The acceleration at a distance 2Rx from the center of mass of Dendor is: gdendor = GMdendor/(2Rx)^2 = G(4Mearth)/(10Rearth)^2 = 4GMearth/100Rearth^2 = 4/100 *(gearth) = (4 · 10 m/s^2)/100 = 0.4 m/s^2

A 10 kg object hangs from a spring balance. The balance reads 5 kg when the object is fully submerged in water, and reads 7.5 kg when the object is fully submerged in an unknown liquid. What is the ratio of the density of this unknown liquid to the density of water?

0.5 Answer is A since: since: From Archimedes' principle buoyancy force Fb = ρgV where volume V stays unchanged since both fully submerged and g is the gravity acceleration. The gravity is 100 N and the buoyant force provided by water is FbWater = 100 N−50 N = 50 N; the buoyant force provided by unknown liquid is FbLiquid = 100 N−75 N = 25 N, which is half of water. So the unknown liquid has half density of water, since g and V are constants. ρliquid = ρwaterFbLiquid/FbW ater = 0.5 · ρwater

A spring with a spring constant of 2 N/m is compressed from equilibrium by a force of magnitude 2 N. How much potential energy is stored by the spring?

1.0 J Answer is A because U = 1/2 kx^2 Solving for x gives x = sqrt(2U/k) |F| = k*sqrt(2U/k) = √ 2U/k Solving for U gives U = |F|^2*1/2k = 1.0 J.

A 20-kg water mill with a radius of 2 m is initially at rest. A constant torque τ = 4.0 N·m is applied continuously for 10 s, making the mill rotate about its axle. What is the magnitude of the angular velocity of the mill at the end of the 10 s period? The water mill should be taken as a solid disc with a radius of 2 m

1.0 rad/s Answer is C since: For constant angular acceleration ω = αt + ω0 I = MR^2/2 = 20 kg × (2 m)^2 /2 = 40 kg · m^2 Initial angular velocity, ω0 = 0, so ω = αt Solve for α using τ = Iα, which gives α = τ /I Therefore ω = αt = τ t/I = 4 N · m(10 s)/40 kg · m^2 = 1 rad/s

On the surface of the Earth an object needs to accelerate to a speed of 11 km/s in order to reach escape velocity. On a planet with 160 times the mass of the Earth, the escape velocity at the surface is 44 km/s. What is the radius of this planet in units of Earth radii?

10 Rearth Answer is E since: The escape velocity is given by vescape = sqrt(2GM/R) Solving for the radius of a planet in terms of its escape velocity: R = 2GM/vescape^2 The radius of the Earth can be written Rearth = 2GMearth/vescape,earth^2 Note that vescape,P /vescape,E = (44 km/s)/(11 km/s) = 4 so vescape,P = 4vescape,E The radius of this planet is thus Rplanet = 2GMplanet/v^2 escape,P = 2G(160Mearth)/(4vescape,earth)^2 = 10 · 2GMearth/vearth^2 = 10Rearth

A spring, which hangs vertically, has a natural angular frequency of 10 rad/s when a block of mass M is attached to the end of the spring. How much is the spring elongated when the mass is at rest?

10 cm Answer is A since: the oscillation equation is d^2x/dt^2 + ω^2x = 0; the solution is x(t) = A · cos(ωt + φ) A free body diagram of the spring and mass gives a balanced force equation kx = mg We can rearrange this to sqrt(k/m) = sqrt(g/x) so the natural angular frequency can be expressed as ω = sqrt(k/m) = sqrt(g/x) = 10 rad/s = sqrt(10 m/s^2/x) x = 10m/s^2 100 rad/s^2 = .1 m = 10 cm

A baseball of mass 145 g is hit exactly horizontally at a height of 1.25 m above ground. Its initial horizontal velocity is 20 m/s. What is approximately the horizontal distance from where the ball was hit to where it touched the ground? (remember, assume g = 10 m/s 2 )

10.0 m Answer is D since: the time of flight is determined by the vertical fall h = (1/2)gt^2 → t = sqrt(2h/g) = sqrt((2*1.25 m)/(10 m/s^2)) = sqrt(0.25 s^2) = 0.5 s The distance is determined from the time of flight and horizontal velocity d = vt = 20 m/s * 0.5 s = 10 m The mass is irrelevant here (i.e., this motion depends on g but not on m)

A U-tube contains two liquids that do not mix. The denser of the two liquids has a density of 1600 kg/m^3 . Using the information contained in the figure, what is the density of the less dense liquid?

1000 kg/m^3 Answer is E since: According to the hydrostatic equilibrium with gravity the liquid pressure under depth h is p = p0 + ρgh The pressure are the same for both liquid cylinders: p1 = p0 + ρ1gh1 = p2 = p0 + ρ2gh2 Solving for density of the less dense liquid which is in the left half part of the U-tube, we can get ρ1 = ρ2h2/h1 = 1600 kg/m^3× 5 cm/8 cm = 1000 kg/m^3

A hydraulic press has one piston of diameter 4.0 cm and a second, larger piston of unknown diameter. The press is balanced when a package with mass of 10 kg is placed onto the smaller piston and a package with mass of 160 kg is placed onto the larger piston. What is the diameter of the larger piston then?

16 cm Answer is E since: From Pascal's law Pressure = Force/Area are the same for both pistons; Rearrange it as Fl/Fs = Al/As = dl^2/ds^2 , Where dl means the diameter of the larger piston and ds the diameter of the smaller piston, so we have dl = ds · p Fl/Fs= 4cm · p 1600N/100N = 4 cm· 4 = 16 cm

A uniform disc of radius Rs = 1 m rotates with angular velocity 4 rad/s about an axis which passes through its center and which is perpendicular to it. Due to internal forces, the mass of the disc is pushed radially outward, turning it into a hoop of the same radius and mass. The magnitude of its angular velocity is now:

2 rad/s Answer is C. Angular momentum is conserved, meaning Ld = Lh or Idωd = Ihωh Thus, ωh = Idωd/Ih Id = 1/2*MRd^2 and Ih = MRh^2 Id/Ih = 1/2*MRd^2/MRh^2 = (1/2), since the masses are the same ωh = 1/2*ωd = 1/2 *4rad/s = 2 rad/s

A tennis ball launcher, which is sitting on a perfectly flat field, shoots a ball into the air at an angle of 30 degrees with respect to the horizontal direction. Ignore air resistance. The initial speed of the tennis ball is 50 m/s. In your calculations remember to take g = 10 m/s^2 About how far from the launcher will the ball hit the ground?

220 m Answer is C since the range R is: R = v0x*T = v0x*2t = (v0*cos α)*(2*v0*sin α/g) = [(2*v0^2)/g]*cos α*sin α = (v0^2/g)*sin 2α R = [(50^2 m^2/s^2)/(10 m/s^2)]*sin(2*30◦ ) = 250*sqrt(3/2 m) ≈ 250*0.866 m = 216 m

A track and field runner achieves a speed of 7.0 m/s. This speed is approximately

25 km/hr Answer is C since: 7 m/s = (7 m/s)*(1 km)/(1,000 m)*(3,600 s)/(1 hr) = (25200 km/(1,000 hr) = 25.2 [km/hr]

A speaker blasting a Ti note (with a frequency of about 500 Hz) moves away from a stationary microphone with a speed of 340 m/s. Recall that the speed of sound in the air is 340 m/s. The frequency of the sound recorded by the microphone is about

250 Hz Answer is A since: Doppler's effect for moving source is f' = f*(u/(u ± vS)) We choose positive sign for receding; f' = f*(u/(u + vS)) = 500 Hz · (340m/s / (340m/s + 340m/s)) = 500 Hz×0.5 = 250 Hz

A pulley (which can be considered to be a solid, massive, uniform cylinder) with mass M = 40 kg is suspended from the ceiling. A massless rope passes over it with a 25.0 kg block attached to one end and a 15.0 kg block attached to the other. The rope does not slip on the pulley. When the speed of the heavier block is 3.0 m/s, what is the total kinetic energy of the pulley and blocks? (Hint: The radius of the pulley is not specified because it is not needed. For the pulley, K = 1/2 Iω^2 ; both I and ω depend on the radius R of the pulley in such a way that it cancels out.)

270 J Answer is C since: K = Ktrans + Krot. The blocks are connected and move with the same speed, so Ktrans = 1/2m1v^2 + 1/2m2v^2 Krot = 1/2 Iω^2 . Since the rope does not slip, the pulley rotates at a speed ω = v/r I of the pulley is 1/2Mr^2 . The rotational kinetic energy is then Krot = 1/2 Iω^2 = 1/2 (1/2Mr^2)(v/r)^2 = 1/4Mv^2 The total kinetic energy is K = 1/2m1v^2 + 1/2m2v^2 + 1/4Mv^2 = 1/2 (m1 + m2 + 1/2M)v^2 = 1/2 (25 kg + 15 kg + 1/2*40 kg)(3 m/s)^2 = 270 J

A uniform steel ball is initially placed on top of a ramp, at an unknown height. It is set free to roll down the ramp without slipping. At the bottom end of the ramp, the translational speed of the ball is 20 m/s. What is the initial height of the ball on the ramp. (Use g = 10 m/s^2)

28 m Answer is D since: When the ball is at the end of the ramp, all of its potential energy has been converted to kinetic energy, meaning K = mgh We need to consider both the linear movement and the rotation K = 1/2mv^2 + 1/2*Iω^2 = 1/2mv^2 + 1/2*I(v/r)^2 For a solid sphere, the rotational inertia is Isphere = 2/5mr^2 leading to K = 1/2mv^2 + 1/5mv^2 = 7/10mv^2 We have mgh = 7/10mv^2 → h = 7/10*v^2/g = 7/10*(20 m/s)^2/10 m/s^2 = 28 m

On a rope, the wavelength of a traveling wave is three meters. If the speed of the waves is 12 m/s, what is the wave number?

2π/3 rad/m Answer is A since: According to the relation: wave number = 2π/λ; And so k = 2π/(3 m)

A wave is described by y(x, t) = A cos(kx − ωt), where x and y are in meters, t is in seconds, A = 0.1 m, k = 2 rad·m^−1 , and ω = 8 rad·s^−1 . The velocity of this wave is

4 m/s Answer is C since: v = ω/k = (8 rad·s^−1)/(2 rad·m^−1) = 4 m/s.

Water flows through a cylindrical pipe of varying cross-section. The velocity is 4.0 m/s at a point where the pipe's diameter is 2.0 cm. What is the diameter of the pipe at a point in which the velocity is 1.0 m/s?

4.0 cm Answer is D since: Water is an incompressible fluid, so A1v1 = A2v2, according to the continuity equation. Thus A2 = A1v1/v2 = 4A1; as for diameter d2/d1 = p A2/A1 = 2. So d2 = 2d1 = 2 × 2.0 cm = 4.0 cm.

A simple pendulum consists of a 15.0 kg mass attached to a massless string. It is released from rest at position ∆y = 80.0 cm above the bottom equilibrium point P, as shown. Its speed at this lowest point P is (recall that 1 cm = 0.01 m)

4.00 m/s Answer is C since Ei = Ui + Ki = mg∆y + 0 as it begins from rest. ∆y = 80.0 cm = 0.80 m Ef = 1/2mv^2 Ei = Ef as energy is conserved, thus mg∆y = 1/2mv^2 The mass cancels. Solving for v gives v = √ 2*g∆y = sqrt(2 · 10 m/s^2 · 0.80 m) = 4 m/s

On a planet different from Earth, two blocks are connected by a string passing over a pulley. The mass of block A is 5.0 kg, and the mass of block B is 15.0 kg. The blocks are moving with an acceleration of 3.0 m/s^2 . Assuming that the string and the pulley are massless, and ignoring any friction or air drag, what is the gravitational acceleration on this planet? Referring to the situation in the previous problem, what is approximately the magnitude of the tension in the string?

45 N Answer is E since using the solution of the previous problem we can take either equation for forces, thus (taking the first one): T − mAg = +mAa T = mAg + mAa = mA(a + g) = 5.0[kg]*(3.0 [m/s^2 ] + 6 [m/s*2 ]) = 45 [N]

A 2.0 kg block is held at some height hi above the ground. It is then released and is falling towards the ground. When its speed reaches 10 m/s, what distance has it fallen (take g = 10 m/s^2)

5 m Answer is B since: Ei = Ef , as energy is conserved for gravitational forces. Ei = Ui + Ki. Initially, Ki = 0, as the block starts at rest. Thus, Ei = mghi Similarly, Ef = Uf + Kf We have kinetic energy as well as potential energy, Ef = Uf + Kf = mghf + 1/2mvf^2 Setting Ei = Ef gives mghi = mghf + 1/2mvf^2 Solving for hi − hf (vertical distance traveled), gives hi − hf = (1/2 mvf^2)/(mg) , so m cancels Plugging in the numbers gives hi − hf = (10 m/s)^2 /(2 · 10 m/s^2 ) = 5 m

A tennis ball launcher, which is sitting on a perfectly flat field, shoots a ball into the air at an angle of 30 degrees with respect to the horizontal direction. Ignore air resistance. The initial speed of the tennis ball is 50 m/s. In your calculations remember to take g = 10 m/s^2 About how long will the ball be in the air before hit the ground?

5 sec Answer is C since: the times to rise and fall are the same and equal to t = v0y/g (from v0y = gt) Note also that vy = 0 at the top of the trajectory T = 2*t = 2*v0y/g = 2*(v0*sin α/g) = (2*50 m/s*sin 30◦)/(10 m/s^2) = (100 m/s*0.5)/(10 m/s^2) = 10*0.5 s = 5 s

A person does 2.5 Joules worth of work by stretching a rubber band from x = 0 to x = 10 cm. When the rubber band is stretched a distance x from the equilibrium, it exerts a restoring force of magnitude F = Ax, where A is a constant. Find the value of A. (Recall, 1 J = 1 N · m.)

5.0 [ N cm ] Answer is D since: A person is pulling the band with a force Fperson = F The displacement is in the same direction, so the angle between the force and the displacement is 0 along the displacement from 0 to d = 10 cm (Recall that cos 0 = 1) Work by a person is W = integral(Fperson(x) · dx) = integral([Fperson(x) cos 0]dx) Since F = Ax, then W = integral((Ax)dx) = 1/2Ax^2 So A = 2*W/d^2 = 2·250 [N·cm]/10^2 [cm^2] = 5.0 [ N/cm ]

A kid is applying a 300 N force in the positive x-direction to a 30 kg desk by pushing it. His dad is pulling the same desk with a 350 N force in the negative x-direction. There is also a horizontal frictional force. If there are no other horizontal forces acting on the system, and the desk is moving at a constant velocity, what is the frictional force?

50 N, in the positive x-direction Answer is C since: the desk is moving with constant velocity so the net force acting on it is 0; the two applied forces, F+x (in the positive x direction) and F−x (in the negative x direction), AND the force of friction fk must add up to 0: F+x = +300 N and F−x = −350 N so: F+x + F−x + fk = 0 thus fk = −(F+x + F−x) = −[300 N + (−350 N)] = +50 N (positive direction!)

An organ pipe measuring 50 cm is open at one end and is installed in room filled with helium. If the speed of sound in helium is 1000 m/s, what is the fundamental frequency of the pipe?

500 Hz Answer is D since: The pipe is open at one end so has one node. The length of the pipe then has to be L = λ/4 to cause resonance in the air column inside, where λ is the wavelength of the fundamental mode. So wavelength λ = 4L = 4 × 50 cm= 2.0 m. Using the relation speed= wavelength × frequency, we solve for frequency, giving f = v/λ = 1000 m s /2.0 m= 500 Hz

A man stands halfway up a 5.0 m ladder of negligible weight. The base of the ladder is 3.0 m from the wall as shown. The magnitude of the force of friction between the ground and the ladder is 210 N. What is the man's weight? Assume that the wall-ladder contact is frictionless

560 N Answer is B since: The friction of the ground on the ladder equals the force the wall exerts on the ladder so ΣFnet = 0 Notice that the L0 = 5 m ladder makes contact with the wall L1 = sqrt((5m)^2 − (3m)^2) = 4 m from the floor, since the distance from the wall to the bottom of the ladder is L2 = 3 m. Consider the net torque about the point of contact of the ladder with the floor, r⊥F = L1 and r⊥W = L2/2 Στnet = W r⊥W − Fwallr⊥F = W(L2/2) − FwallL1 = 0 Note also that Fwall = ffriction = 210 N Then Fwall = (L2/2L1)W solving for W gives 560 N

A truck with a mass of 2,000 kg is about to drive through a curve with a radius of 600 m. The largest force of friction attainable between the truck's tires and the road is 3, 000 N. With what maximum momentum can the truck traverse this bend without slipping?

6 × 104 kg*m/s Answer is A since: In this problem for a uniform circular motion the truck of mass m = 2, 000 kg on a curve has centripetal acceleration a = v^2/R, where R = 600 m and the velocity v is unknown The largest force of friction the truck's tires can "hold up" is fs = 3, 000 N This fs "provides" the centripetal force in this circular motion: Fr = mv^2/R fs = Fr = ma*r = mv^2/R; The question is about the maximum momentum We recall that p = mv thus mv^2 = (m/m)(mv^2) = p^2/m Thus, fs = (p^2/m*R) → p = sqrt(fs*m*R) = sqrt(3, 000 [N]*2, 000 [kg]*600 [m]) = sqrt(3 × 103 × 2 × 103 × 6 × 102[N*kg*m]) = sqrt(36 × 108[( kg*m/s^2) *kg*m] = 6 × 104 kg*m/s

On a planet different from Earth, two blocks are connected by a string passing over a pulley. The mass of block A is 5.0 kg, and the mass of block B is 15.0 kg. The blocks are moving with an acceleration of 3.0 m/s^2 . Assuming that the string and the pulley are massless, and ignoring any friction or air drag, what is the gravitational acceleration on this planet?

6.0 m/s^2 Answer is E since: Both masses, mA and mB, are moving with the same acceleration (smaller mass mA moving up, larger mass mB moving down) Choose a (conventional) coordinate system with the y axis pointing up From the Newton's 3rd Law, the force of tension T acting on each mass is identical and acting in the opposite direction. There are two forces acting on each mass: tension T and gravity g, thus: P(A) = T − mAg = +mAa P(B) = T − mBg = −mBa Solve for g subtracting (on both sides) the 2nd equation from the 1st: (T − mAg) − (T − mBg) = (+mAa) − (−mBa) T − T − mAg + mBg = +mAa + mBa (mB − mA)g = (mB + mA)a thus g = mB+mA mB−mA a = (15.0[kg]+5.0[kg])/(15.0[kg]−5.0[kg])*3.0 [m/s^2] = 6.0 m/s^2

A 0.5 kg block is held in place compressing an ideal spring by a distance xm = 5.0 m from its point of equilibrium. The block is then released and as a result can slide up a frictionless ramp, as shown in the figure. If the spring constant is 3 N/m, what height h will the block reach? In your calculations, ignore the size of the block

7.50 m Answer is D since Ei = 1/2 kx^2 = Ef = mgh Solving for h gives h = (kx^2)/(2mg) = (3 N/m · (5 m)^2)/(2 · 0.5 kg · 10 m/s^2) = 7.5 m

A uniform meter stick weighing 200 grams can be balanced horizontally with a 300 g mass placed at the 100 cm mark (i.e., at one of its ends) if a fulcrum underneath (the point on which a lever is supported and on which it pivots) is placed at the point marked:

80 cm Answer is E since: Στi = 0 The torque exerted by the weight is mg · (L100 − d), where d is the location of the fulcrum and L100 the 100 cm mark. The torque exerted by the meter stick's weight is Mg · (L50 − d) where L50 is the 50 cm mark. They must sum to zero, so Στnet = Mg(L50 − d)) +mg(L100 − d) = 0 Solving for d we get d = (L100 · m + L50 · M)/(M + m) = 100cm×600g+50cm×400g 400g+600g = 80 cm

The vector A is given by: A = −3ˆi + 5ˆj + 8ˆk and the vector B is given by: B = 1ˆi − 3ˆj + 9ˆk. What is the magnitude of the vector C, where: C = A − B?

9 Answer is D since: |C| = sqrt[(Ax − Bx)^2 + (Ay − By)^2 + (Az − Bz)^2] = sqrt[(−3 − 1)^2 + [5 − (−3)]^2 + (8 − 9)^2 = sqrt[(−4)^2 + 8^2 + (−1)^2 = sqrt[16 + 64 + 1] = sqrt(81) = 9

A uniform rigid rod AB is supported at points X and Y, as shown. The rod AB has length of L and weight of 350 N. The distance from end A to X is L/4 and the distance from end B to Y is L/6. The supporting forces at X and Y are

FX = 200 N, FY = 150 N Answer is B since: The sum of forces ΣFnet = FX + FY − W = 0; so FX + FY = W = 350 N Choosing the middle of the rod O as the pivot, XO = L/2 − L/4 = L/4, YO = L/2 − L/6 = L/3; and OX⊥FX, OY⊥FY , OO = 0 The sum of torques Στnet = FX× OX + FY × OY −W× OO = FX × L/4 + FY × L/3 − W × 0 = 0; so FY = 3FX/4 Solving the two equations about FX and FY above, we get FX = 4W/7 = 200 N, FY = 3W/7 = 150 N

In the figure above, IA, IB and IC are respectively the rotational inertia of the rotating systems (A), (B) and (C). Which of the following statements is true?

IA = IC not equal to IB Answer is C. We have I = Σmiri^2 IA = (1 × 1)^2 + (1 × 1)^2 + (2 × 0)^2 = 2 kg m^2 IB = (1.5 × 1)^2 + (1.5 × 1)^2 = 3 kg m^2 IC = (0.5 × 2)^2 = 2 kg m^2

A block of mass 5 kg sits motionless on an inclined plane. The inclined plane has an inclination of 30◦ with respect to the horizontal plane. What is approximately the magnitude of the static frictional force on the block?

fs = 25.0 N Answer is C since: The body is not moving so the force due to gravity pushing it down the incline plane is counter-acted by a force of friction (i.e., the two forces are equal in magnitude and opposite in direction): F|| − fs = 0 → F|| = fs The component of gravitational force acting along the incline is F|| = mg sin θ = 5 [kg]*10 [m/s^2 ]*sin 30◦ = 50*0.5 [N] = 25.0 [N].

Three identical rods resting on a frictionless table are acted on by forces of equal magnitude, represented by arrows in the figure below. (The figure should be interpreted as a view from above the table.) Which of the three rods are in static equilibrium? (Ignore the force of gravity and other outside forces)

none of them Answer is E since: The sum of forces ΣF = 0 for equilibrium; so neither rod#1 or rod #3 is in static equilibrium Similarly, the sum of torques Στ = 0 for equilibrium; so rod #2 cannot be in static equilibrium In conclusion, none of them are in static equilibrium

A top spinning on the floor precesses because of the torque due to gravity about the point of contact of the top with the floor is causing the angular momentum of the top to change. The direction of this torque is

parallel to the floor Answer is E since: Since τ = r × F, torque acts in a direction perpendicular to both the radius and the force. The force in this case is gravity, which points down, while the radius points along the axis of rotation. Using the right hand rule gives a direction parallel to the floor

The position x of an object has the dimension of [L] (i.e., length). Its velocity v has the dimension of [L]/[T] (i.e., length/time) and its acceleration a has the dimension of [L]/[T]2 . Which of the following quantities has the same dimension as v/a ?

x/v Answer is A since DIM(x/v) = [L]/([L]/[T ]) = [T]

What is the wavelength of a string fixed at both ends?

λ = 2L/m

Two particles, each of mass 4M, are a distance d apart. To bring a third particle, with mass M, from far away to a point midway between the two particles, the work done by an external agent is given by:

−16GM^2/d Answer is E because: W = ∆U = Uf − Ui. The total potential is the sum of the gravitational potential between each object Ui = −G*(m1m2)/(r1,2) = −G (4M)^2/d Uf = −G*(m1m2)/(r1,2) − G*(m1m3)/(r1,3) − G*(m2m3)/(r2,3) = −G (4M)^2/d − G (4M)M/(d/2) − G (4M)M/(d/2) Here m1 and m2 are the initial particles, and m3 is the one brought in. The potential between particle 1 and 2 does not change, so the change in potential is U1,3 + U2,3 = −G (4M)M/(d/2) − G (4M)M/(d/2) = −G 8M^2/d − G 8M^2/d = −16G M^2/d

As a box with a mass of 4.0 kg is transported 4.5 m vertically upward, what is the work done by gravity?

−180 J Answer is B since: We can choose a coordinate system with y axis pointing up Force of gravity Fg is acting in the negative vertical direction. Force lifting the block is acting in the positive direction and the displacement d (vector!) is in the positive direction. The angle between vectors of the force of gravity and the displacement is π (or 180◦ ) So the work done by the gravitational force is W = Fg · d = Fg · d cos π = mg*d cos π = 4.0 [kg] · 10 [m/s 2 ] · 4.5 [m](−1) = −180 [J]