Problem 1:
(a) An astronaut on a strange
planet finds that she can jump a maximum horizontal distance of 15 m if her
initial speed is 3 m/s. What is the free-fall acceleration on the planet?
(b) How much work is required to raise a 100 g block to a
height of 200 cm and simultaneously give it a velocity of 300 cm/sec?
Solution:
 | Concepts, principles, relations
that apply to the problem: (a) Projectile motion, (b) Energy and work |
| Why do they apply? (a) After she has left the ground, the astronaut becomes a projectile. (Her motion is motion in more than one dimension with constant acceleration.) The maximum range of a projectile is associated with a launch angle of 45o. (b) To change the total energy of an object, an external force must do work. |
| How do they apply? (a) range: R = v02sin2q0/g’, Rmax = v02/g’, g’ = (9/15)(m/s2) = 0.6 m/s2. (b) W = mgh + ½mv2 = 0.1*9.8*2 J + 0.5*0.1*9 J = 2.41 J. |
| Details of the calculation: None |
Problem 2:
A 36-g bullet with a speed of 350 m/s strikes a 8-cm thick
fence post. The bullet is retarded by an average force of 3.6*103 N
while traveling all the way through the board.
(a) What speed does the bullet have when it emerges?
(b)
How many such boards could the bullet penetrate?
Solution:
| Concepts, principles, relations that apply to the
problem: Newton's second law, F = ma, motion with constant acceleration in one dimension |
| Why do they apply? For motion with constant acceleration the equations of kinematics yield vxf2 = vxi2 + 2ax(xf - xi). |
| How do they apply? (a) (xf - xi) = 0.08 m, vxf2 = (350 m/s)2 - 2*(0.08 m)(3.6*103 N)/( 36*10-3 kg), vxf = 326.34 m/s. (b) vxf = 0, (350 m/s)2 = 2*d(3.6*103 N)/( 36*10-3 kg), d = 0.6125 m. # of boards the bullet can penetrate = 7, (d/0.08 = 7.65) |
| Details of the calculation: None |
Problem 3:
A highway curve with a radius of 750 m is banked properly for a car traveling 120 km/h. If a 1590 kg Porsche rounds the curve at 230 km/h,
| (a) how much sideway force must the tires exert against the road if the car does not skid? |
| (b) what must be the bank angle for the Porsche to turn if there is no friction force on the tires? |
Solution:
| Concepts, principles, relations
that apply to the problem: Uniform circular motion, Fc = mv2/r |
| Why do they apply? In part (a) the frictional force must provide part of the centripetal acceleration, in part (b) the horizontal component of the normal force must provide all the centripetal acceleration. |
| How do they apply? (a) A free-body diagram of the car on the track is shown below.  When the car is traveling at v = 120 km/h the frictional force f = 0 and Ncosq = mg, Nsinq = mv2/r, tanq = v2/(gr), q = 8.6o. When the car is traveling at v' = 230 km/h we have Ncosq - mg - fsinq = 0, Nsinq + fcosq = mv'2/r. f(tanqsinq + cosq) = mv'2/r - mgtanq. f = 6226 N. The tires must exert a sideway force of 6226 N. |
| Details of the calculation: For part (b) the horizontal component of the normal force must equal mv'2/r = 8653.4 N. The vertical component of the normal force is mg = 1590 kg(9.8 m/s2) = 15582 N tanq = 8653.4/15582 q = 29o. The bank angle must be 29o. |
Problem 4:
A 20-kg child sits on a turntable at a distance of 1.2 m
from the center. The coefficient of static friction between the child and the
turntable is 0.6.
(a) If the turntable is rotating at a frequency of 3 revolutions per minute,
what is the frictional force exerted by the platform on the child?
(b) At what frequency of rotation will the child slide off the platform?
Solution:
| Concepts, principles, relations
that apply to the problem: Centripetal acceleration, the frictional force |
| Why do they apply? The frictional force has to provide the centripetal acceleration. |
| How do they apply? (a) fs = mv2/r, m = 20 kg, r = 1.2 m, v = 2pr/(20 s). fs = 2.37 N. (b) fs(max) = 0.6*20 kg*9.8 m/s2 = 117.6 N, v(max) = (fs(max)*r/m)½ = 2.66 m/s. frequency: f = v(max)/(2pr) = 0.352 revolutions /s. |
| Details of the calculation: None |
Problem 5:
A mass is suspended from a fixed point by a light cord of length L. The mass is set in motion in a horizontal circle of radius R as shown. Such an arrangement is called a conical pendulum because the moving cord sweeps out the surface of an inverted cone.
| (a) What is the frequency f of such a conical pendulum in terms of L, R, and g? |
| (b) Show that for L much greater than R the period (T) of a simple pendulum is approximately the same as that of a conical pendulum of the same length. |
Solution:
| Concepts, principles, relations
that apply to the problem: Uniform circular motion, Fc = mv2/r |
| Why do they apply? The mass moves along a circular path with constant speed v. The centripetal acceleration is v2/R. The magnitude of the horizontal component of the tension must provide the centripetal force Fc = mv2/R, while the magnitude of the vertical component of the tension must be equal to weight mg. |
| How do they apply? (a) (mv2/R) /[(mv2/R)2 + (mg)2]1/2 = R/L, v4 = R4g2/(L2-R2). The period of the pendulum is T = 2pR/v = 2p(L2-R2)1/4/g1/2. The frequency is f = 1/T. (b) If L >> R then T @ 2p(L/g)1/2. This is the period of a simple pendulum of length L. |
| Details of the calculation: None |
Problem 6:
A ball is weighing 4 pounds is suspended by a weightless string of length 6 ft. The ball is delivered a horizontal impulse of magnitude 3 pounds-seconds. Find the maximum height above its initial position, attained by the ball. Take 32 ft/s2 as the gravitational acceleration and assume there is no friction at the point of attachment of the string.
Solution:
| Concepts, principles, relations
that apply to the problem: This is a problem with constraints, so working directly from Newton’s laws is not possible. We will use conservation laws. |
| Why do they apply? We treat the ball as a point particle. (Not enough information is given to treat it as a rigid body.) Initially the ball receives an impulse of magnitude FDt = p0. It’s kinetic energy is now p02/2m. It will start moving in a circle. The forces on the ball are the force of gravity and the tension in the string. If the initial impulse is large enough so that the ball rises above the pivot point, then the centripetal acceleration needed to keep the ball moving on the circle is partly provided by gravity and the tension in the string decreases. When the tension in the string vanishes then the ball behaves like a free particle in a gravitational field. |
| How do they apply? Refer to the figure below:  Energy conservation: E = ½p02/(2m) = ½mv2 + mg (l + lsinq'), v2/l = 2E/(ml) - 2g(1+sinq') is the centripetal acceleration. The tension in the string vanishes when v2/l = g sinq', i.e. when all the centripetal acceleration is due to gravity. Then sinq' = 2E/(3mgl) - 2/3. |
| Details of the calculation:E = p02/(2m)
= (9 lb2s232 ft)/(8 lbs2) = 36 lb ft,
since m = 4 lbs2/(32 ft) . l = 6 ft. Therefore sinq’ = 1/3, q’ = 19.47o. At q’ = 19.47o we have v = (lgsinq’)½, vx = vsinq’, vy = vcosq’. The ball now behaves like a projectile. It will gain an additional height Dh. ½mvy2 = mgDh, Dh = (lsinq’cos2q’)/2. Dh = 3 ft 0.296 = 0.89 ft Maximum height: l + lsinq’ + Dh = 6 ft + 2 ft + 0.89 ft = 8.89 ft |
Problem 7:
You are an astronaut, and you observe a small planet to be spherical. After landing on the planet, you set off, walk straight ahead, and find yourself returning to your spaceship from the opposite side after completing a lap of 25 km. You hold a hammer and a falcon feather at a height of 1.4 m, release them, and observe them to fall together to the surface in 29.2 s. What is the mass of the planet?
Solution:
| Concepts, principles, relations
that apply to the problem: Newton’s law of gravitation, one dimensional motion with constant acceleration |
| Why do they apply? Near the surface of the planet the gravitational acceleration is constant, pointing downward. Newton's law of gravitation, F = GMm/R2 lets us calculate the magnitude of this acceleration as a function of the mass M and the radius R of the planet. Walking around the planet yields its circumference, and given the circumference of the planet, therefore the radius R can be determined. The acceleration g is measured by dropping the objects through a known distance and measuring the time interval Dt. The objects fall together, so we know there is no drag due to an atmosphere. |
| How do they apply? Dy = ½gDt2 yields g = 2*1.4 m/(29.2 s)2, 2pR = 25 km, yields R = 25 km/2p |
| Details of the calculation: g = GM/R2, M = gR2/G = [2*1.4 m/(29.2 s)2 * (25 km/2p)2/(6.67 10-11m3kg-1s-2) = 7.8 1014 kg |
Problem 8:
Assuming that the earth is a uniformly dense sphere and that the acceleration of gravity is g at the surface, what is the acceleration of gravity below the surface as a function of distance from the center?
Solution:
| Concepts, principles, relations
that apply to the problem: Newton’s law of gravitation |
| Why do they apply? The gravitational acceleration a distance r from the center is in the –r direction and its magnitude is found using “Gauss’ law”. |
| How do they apply? Imagine a tunnel bored through the center of the earth. For a test particle a distance r from the center we have 
4pGMinsideAt the surface g = (4/3)GprR, so a = gr/R, where R is the radius of the earth. |
| Details of the calculation: None |
Problem 9:
A particle of mass m is given an initial velocity v0i.
Assume that the particle is subject to a drag force F = -bv½i.
(a) Find v as a function of time.
(b) How far does the particle travel before coming to rest?
Solution:
| Concepts, principles, relations
that apply to the problem: Newton's second law |
| Why do they apply? Given the acceleration and v(0) we can find v(t) and d(t) as a function of time. |
| How do they apply? (a) mdv/dt = -bv½, dv/dt = -(b/m)v½, dv/ v½ = -(b/m )t, òv0v(t) dv/v½ = -(b/m) ò0tdt = -bt/m, 2(v(t)½ - v0½) = -bt/m, v(t) = (v0½ - bt/(2m))2. (b) v(t) = 0 à tf = 2mv0½/b. After a time interval tf the particle stops. In the time interval tf the particle travels a distance Dx. Dx = ò0tfv(t) dt = ò0tf(v0½ - bt/(2m))2dt = [(v0½ - btf/(2m))3- v03/2]/(-3b/(2m)) Dx = 2mv03/2/(3b). |
| Details of the calculation: None |
Problem 10:
A mass M is attached to the end of a string which passes through a hole in a frictionless horizontal plane. Initially the mass moves on a circle of radius R with kinetic energy T0. The string is then slowly pulled so that the mass finally rotates on a circle of radius R/3. How much work was done?
Solution:
| Concepts, principles, relations
that apply to the problem: Conservation of angular momentum |
| Why do they apply? We have a force directed towards the center of the orbit. Therefore the torque is zero. |
| How do they apply? Initial angular momentum: Li = mRv0 = mR(2T0/m)1/2. Final angular momentum: Lf = mRvf /3 = m(R/3)(2Tf/m)1/2. Conservation of angular momentum: Li = Lf, mR(2T0/m)1/2 = m(R/3)(2Tf/m)1/2, T0 = Tf/9, Tf = 9 T0. Work done: Tf – T0 = 8 T0. |
| Details of the calculation: None |
Problem 11:
A ball that is thrown upward near the surface of the earth with a velocity of 50 m/s will come to rest about 5 second later. If the ball were thrown up with the same initial velocity on Planet X, after 5 seconds it would still be moving upwards at nearly 31 m/s. The magnitude of the gravitational field near the surface of Planet X is what fraction of the gravitational field near the surface of the earth?
Solution:
| Concepts, principles, relations that apply to the
problem: One-dimensional motion with constant acceleration |
| Why do they apply? We are given the velocity at t = 0 and t = 5s and are asked to find the acceleration. |
| How do they apply? v(t) = v0 + at = v0 - g't. 31m/s = 50m/s - g'(5s). g' = 3.8m/s2. g'/g = 0.388. |
| Details of the calculation: None |
Problem 12:
Write F = ma in a rotating coordinate system and identify Coriolis and Centrifugal terms.
Solution:
| Concepts, principles, relations
that apply to the problem: Motion in a non-inertial frame |
| Why do they apply? A rotating coordinate system is a non-inertial frame. Fictitious forces appear. |
| How do they apply? The relationship between the time derivatives of any vector A in a fixed, inertial frame and in a frame rotating with constant angular velocity W is (dA/dt)fixed = (dA/dt)rotating + W ´ A. Therefore: (dr/dt)fixed = (dr/dt)rotating + W ´ r. The relationship between the velocity vi in the inertial frame and the velocity v in a frame rotating with constant angular velocity W is vi = v + W ´ r. Differentiating again we have (dvi/dt)fixed = (dv/dt)fixed + W ´ (dr/dt)fixed. Inserting (dr/dt)fixed from above we have m(dvi/dt)fixed = Fi = m(dv/dt)fixed +m W ´ v + mW ´ (W ´ r). Now we use (dv/dt)fixed = (dv/dt)rotating + W ´ v to obtain the equations of motion in the rotating frame. |
| Details of the calculation:The equations of motion in the rotating frame therefore are mdv/dt = Fi - 2mW ´ v - mW ´ (W ´ r). If Fi is derivable from a potential then Fi = -¶U/¶r. 2m(v ´ W) = -2m(W ´ v) = Coriolis force m(W ´ r) ´ W = -m W ´ (W ´ r) = centrifugal force |
Problem 13:
Consider a lawn sprayer consisting of a spherical cap (a0 = 45o) provided with a large number of equal holes through which water is ejected with speed v0. The lawn is not evenly sprayed if the holes are evenly spaced. How must the number of holes per unit area, r(a), be chosen to achieve uniform spraying of a circular area? Assume the radius of the sprinkler cap is very much less than the radius of the area to be sprayed, and the surface of the cap is at the level of the lawn.
Solution:
| Concepts, principles, relations that apply to the
problem: Projectile motion |
| Why do they apply? The drops are projectiles. They execute motion in more than one dimension motion with constant acceleration. This acceleration is directed downward and has magnitude g. |
| How do they apply? A drop sprayed from an angle a travels a distance R(a). We want to find this range R in term of a. We use vy0 = v0cosa, vx0 = v0sina, x = x0 + vx0t, y = y0 + vy0t – ½gt2, and substitute 0 = 0 + v0(cosa)t – (½gt2), t = 2v0cosa /g to solve for R = x(t) = v0(sina)t = v0(sina)2v0(cosa)/g = v02(sin2a )/g. Drops sprayed from any angles between a and a + da fall on an area 2pRdR between R and R + dR. The number of drops sprayed from angles between a and a + da is proportional to r(a)2pr2capsin a da. We want [r(a)2pr2capsina da]/[2pRdR] = const, or (r(a)sina)/(RdR/da) = const. |
| Details of the calculation: dR/da = 2v02(cos2a)/g. Therefore r(a )(sina)/(cos2a sin2a) = const, or r(a ) µ sin(4a) /sina. |