Problems

Problem 1:

A 90kg fullback running east with a speed of 5 m/s is tackled by a 95 kg opponent running north with a speed of 3 m/s.  If the collision is perfectly inelastic, calculate the speed and the direction of the players just after the tackle.

Solution:

Concepts, principles, relations that apply to the problem:
Conservation of momentum
Why do they apply?
If a component of the total external force acting on a system is zero, then the corresponding component of the total momentum is conserved.  We assume that the horizontal component of the external force can be neglected during the collision.
How do they apply?
The initial momentum of player 1 is p1 = (90 kg 5 m/s)i = (450 kgm/s)i.
The initial momentum of player 2 is p2 = (95 kg 3 m/s)j = (285 kgm/s)j.
Momentum is conserved, the final momentum p of both players is p = p1 + p2.
Details of the calculation:
p = (m1 + m2)v.
v = (2.432 m/s)i + (1.54 m/s)j.
v2 = 8.29(m/s)2, v = 2.88 m/s.
The speed of the players after the collision is 2.88 m/s.
tanq = py/px = 285/450 = 0.63, q = 32.34o.
Their direction of travel makes an angle q = 32.34o with the x-axis.  (The x-axis is pointing east.)

Problem 2:

After a completely inelastic collision between two objects of equal mass, each having initial speed v, the two move off together with speed v/3.  What was the angle between their initial directions?

Solution:

Concepts, principles, relations that apply to the problem:
Conservation of momentum
Why do they apply?
The external force acting on the system is zero, the total momentum is conserved.
How do they apply?

p1x + p2x = pfx = pf,  p1y + p2y = 0.
2mvcosq = 2mv/3, cosq = 1/3, q = 70.5o.
The angle between their initial directions is 2q = 141o.
Details of the calculation:
None

Problem 3:

A 3 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60o with the surface.  It bounces off with the same speed and angle.  If the ball is in contact with the wall for 0.2s, what is the average force exerted on the ball by the wall?

Solution:

Concepts, principles, relations that apply to the problem:
Impulse: dp = Fdt,  Dp = FavgDt.
Why do they apply?
In the elastic collision with the "infinitely" massive wall, the ball receives an impulse Dp.  Enough information is given to calculate Dp.  Since the contact time is given, Favg can be calculated.
How do they apply?

The balls initial momentum is
pi = pxii + pyij = (3 kg 10 m/s)sin60oi+(3 kg 10 m/s)cos60oj.
Its final momentum is
pf = pxfi + pyfj = -(3 kg 10 m/s)sin60oi + (3 kg 10 m/s)cos60oj.
Dp = pf - pi= -2(30 kgm/s)sin60oi = -(51.96 kgm/s)i.
Dp = Favg Dt.
Favg = -(51.96 kgm/s)i/(0.2s) = -259.8 Ni.
Details of the calculation:
None

Problem 4:

A car of total mass M1 = M and velocity v1 makes a totally inelastic collision at time t = 0 with a second car of mass M2 = 2M at rest.  Before the collision a point object of mass m << M was sitting at the bottom of a frictionless spherical cavity of radius r embodied inside the first car.  For what range of velocities v1 will the small mass lose contact with the surface of the cavity?

Solution:

Concepts, principles, relations that apply to the problem:
Newton's laws, conservation of momentum, frame transformations
Why do they apply?
Before the collision car 1 and mass m move with uniform velocity v1.  The impulse of the collision changes the velocity of car 1, and after the collision car 1 moves with uniform velocity v2.  The small mass moves with velocity v = v1 - v2 with respect to car 1 immediately after the collision.   After the collision the rest frame of car 1 is an inertial frame and we can analyze the problem in that frame.

How do they apply?
Conservation of momentum:  Mv1 = 3Mv2,  v2 = v1/3, v = v1-v2 = 2v1/3.
In the rest frame of car 1 immediately after the collision we have a particle of mass m at the bottom of a cavity of radius r moving with velocity v.  The particle will start moving in a circle.  The forces on the particle are the force of gravity and the normal force from the wall of the cavity.   If the initial speed is large enough so that the particle gains a height greater than r, then the centripetal acceleration needed to keep the particle moving on the circle is partly provided by gravity and the normal force decreases.  When the normal force vanishes then the particle behaves like a free particle in a gravitational field and looses contact with the wall.

Details of the calculation:
When (1/2)mv2 < mgr,  or  v < (2gr)1/2,  or  v1 < (3/2)(2gr)1/2 the particle will not break contact with the wall.
When the particle reaches the top of the cavity with a speed vtop, so that the centripetal acceleration vtop2/r > g, then it will not break contact with the wall. 
Then (1/2)mv2 = 2mgr + (1/2)mvtop2 > 2mgr + (1/2)mgr = (5/2)mgr.
v > (5gr)1/2,  v1 < (3/2)(5gr)1/2.
For initial velocities (3/2)(2gr)1/2 < v1 < (3/2)(5gr)1/2 the particle breaks contact with the wall.

Problem 5:

A rope of length l slides over the edge of a table.  Initially a piece x0 of it hangs without motion over the side of the table.  Let x be the length of rope hanging vertically at time t.  The rope is assumed to be perfectly flexible.

Show that E = T + U is a constant of motion.

Solution:

Concepts, principles, relations that apply to the problem:
Conservation of mechanical energy
Why do they apply?
We are asked to show that the mechanical energy is conserved in a given situation.
Why do we expect mechanical energy to be conserved in this situation?
The only external forces acting are gravity on the vertical part of the rope and the normal force and gravity on the horizontal part of the rope.  Gravity is a conservative force and the normal force acts perpendicular to the velocity and therefore does no work.
How do they apply?
Let l be the length of the rope and let x be the part hanging over the edge.

This is an effectively one-dimensional problem.  The picture below shows a blowup of the corner, the section of the rope bending around the corner has infinitesimal length.


Fg - T = drxv/dt for the vertical part of the rope, T = dr(l-x)v/dt for the horizontal part of the rope.
Defining p = rlv = rldx/dt we have  Fg = dp/dt =  rld2x/dt2 = rgx.
d2x/dt2 = gx/l.
T = (1/2)rl(dx/dt)2,  U = -rg(x2/2), E = T + U = (1/2)rl(dx/dt)2 - rg(x2/2)
dE/dt = rl(dx/dt)d2x/dt2 - rgxdx/dt = rgxdx/dt - rgxdx/dt = 0.
Details of the calculation:
None

Problem 6:

A chain lies pushed together at the edge of a table, except for a piece which hangs over it, initially at rest.  The links of the chain start moving, one at a time.

Show that E = T + U is not a constant of motion.

Solution:

Concepts, principles, relations that apply to the problem:
Conservation of mechanical energy
Why do they apply?
We are asked to show that the mechanical energy is not conserved in a given situation.
Why do we expect mechanical energy to not be conserved in this situation?

Consider two adjacent links of the chain.  Link 1 (red) moves away from link 2 (blue).  Initially the two links do not interact.  After link 1 has traveled a distance Dx, it interacts with link 2 (blue) and the two links stick together and move together.  The two links "collide" inelastically.  In an inelastic collision mechanical energy is not conserved.
How do they apply?
Let l be the length of the chain that hangs over the edge.  This part of the chain is moving. Let the x-axis point downward and let the tabletop be at x = 0.
px = rlv = rldl/dt.  Fx = dpx/dt =  r(dl/dt)2 +  rld2l/dt2 = rlg.
T = (1/2)rlv2 = (1/2)rl(dl/dt)2,  U = -rlg(l/2),
E = T + U = (rl/2)[(dl/dt)2 - gl].
dE/dt = (r/2)(dl/dt)3 + rl(dl/dt)d2l/dt2 - rgldl/dt.
d2l/dt2 = g - (1/l)(dl/dt)2, from Fx = dpx/dt.
dE/dt = (r/2)(dl/dt)3 + rlg(dl/dt) - r(dl/dt)3- rgldl/dt = -(r/2)(dl/dt)3 = -(r/2)v3 ¹ 0.

Problem 7:

A body of mass M slides down a frictionless inclined plane of angle q, starting from the top.  To the body a chain is attached which is coiled at the top of the incline and for which the weight per unit length is r.  By taking into account the uncoiling of the chain, derive the equation for the velocity of the system as a function of the distance traveled along the incline.

Solution:

Concepts, principles, relations that apply to the problem:
Newton's 2nd law, F = dp/dt
Why do they apply?
Newton's 2nd law, F = dp/dt, yields the equations of motion, which can be solved for the velocity as a function of position.
How do they apply?
Let x be the distance traveled along the incline.  Fx = (rx + Mg)sinq.
Note: r is defined as the weight per unit length and not the mass per unit length in this problem.
Fx = dpx/dt = (d/dt)[((rx/g) + M)v] = (r/g)v2 + ((rx/g) + M)dv/dt.
Details of the calculation:
dv/dt = (dv/dx)(dx/dt) = (dv/dx)v.
dv/dx = (1/v)dv/dt = (Fx - (r/g)v2)/[v((rx/g) + M)] = gsinq/v - (rv)/(rx + Mg).
Let x' = rx + Mg, then dv/dx' = (1/r)dv/dx.
dv/dx' = gsinq/(rv) - v/x' = A /v - v/x' with A = gsinq/r a constant.
vdv/dx' = A - v2/x', dv2/dx' = 2A - 2v2/x'.
Try v2 = C1x' + C2/x'2.  Then dv2/dx' = C1 - 2C2/x'3 = 3C1 - 2v2/x',
since 2v2/x' = 1C1 + 2C2/x'3.
We need 2A = 3C1, C1 = 2A/3.
The initial conditions are v2 = 0 and x' = Mg.
0 = 2AMg/3 + C2/M2g2, C2 = -2AM3g3/3.
Therefore
v2 = 2gsinq(rx + Mg)/(3r) - 2g4M3sinq/[3r(rx + Mg)2]
= [2gsinq/(3r)][(rx + Mg) - g3M3/(rx + Mg)2].

Problem 8:

A chain of mass m and length L rests with (1 - a)L of its length on a table top and aL of its length hanging over the smooth edge, as shown in the following figure.

(a)  The coefficient of friction of the tabletop is m.  What is the maximum value, ac, for which the chain remains stationary?
(b)  If a is larger than ac, when released the chain will slide off the table.  What is the velocity of the chain when the last link leaves the table?

Solution:

Concepts, principles, relations that apply to the problem:
Newton's 2nd law:  F = dp/dt,
Work-kinetic energy theorem: òF×dr = DT.
Why do they apply?
This is an effectively one-dimensional problem.  The forces acting on the chain are gravity, the normal force, and friction.
òF×dr = (1/2)mv2.
How do they apply?
(a)  r = m/L = linear mass density,
racLg - r(1-ac)Lgm = 0,
gravity     frictional force
ac - (1-ac)m = 0,  ac = m/(m+1).
Details of the calculation:
(b) 

Problem 9:

A rocket with initial mass m0 starts from rest and travels in a straight line in a gravity-free environment.  It burns its fuel at a constant rate km0, and exhausts the burned gases at a speed v’ relative to the rocket shell of mass m1.  Find the maximum momentum of the rocket.

Solution:

Concepts, principles, relations that apply to the problem:
Motion in one dimension, Newton's 2nd law, F = dp/dt, dp = p(t+dt) - p(t)
Why do they apply?
The system consists of the rocket body and the fuel.  The speed of a small amount of fuel |dm| decreases by -v' in the time interval dt.  In the same time interval the speed of the rocket and the remaining fuel increases by dv.  The total momentum change of the system in the time interval dt therefore is
dp = p(t+dt) - p(t) = (m-|dm|)(v+dv) + |dm|(v-v') - mv = mdv - |dm|v'.
(Only first order terms are retained.)
By Newton's 2nd law dp/dt is equal to the external force F = 0.
How do they apply?
 0 = m(dv/dt) - |dm/dt|v'.
Here |dm/dt| = km0.  Therefore dv/dt = km0v'/m.
At t = 0 m = m0.
For t > 0 and m > m1 we have m = m0 - km0t.  Therefore
dv/dt = km0v'/(m0 - km0t) = kv'/(1-kt).
At time tc the fuel is used up. 
m1 = m0 - km0tc, tc = (1-m1/m0)/k.
For t < tc, dv/dt = - kv'/(1-kt).
Details of the calculation:
Let t' = 1 - kt.  Then dv/dt' = -v'/t'. 
v(t’) = -v’ln(t’/t0’).
v(t) = -v’ln(1-kt).
The momentum of the rocket at time t is p = m0v’(kt - 1)ln(1 - kt )
The function (kt - 1)ln(1 - kt ) has a maximum when ln(1 - kt) = -1, kt = 1 - e-1 = 0.632.
So if ktc = (1-m1/m0) < 0.632, then the maximum momentum of the rocket is
p = -m1v’ln(m1/m0) at time t = tc, otherwise the maximum momentum of the rocket (shell and remaining fuel) is p = 0.368m0v’ at time t = 0.632/k.

Problem 10:

A spherical water droplet falls without friction under the influence of gravity through an atmosphere saturated with water vapor.  Let its initial radius at t = 0 be C, its initial velocity v0.  As a result of condensation, the water drop experiences a continuous increase in mass, proportional to its surface.  Its radius r then increases linearly with time.  Integrate the differential equation of the motion by introducing r instead of t as independent variable.  Show that for C = 0 the velocity increases linearly with time.

Solution:

Concepts, principles, relations that apply to the problem:
Newton's 2nd law:  F = dp/dt, dp = mdv + vdm
Why do they apply?
The system consists of the moisture, with zero momentum. and the water droplet with momentum p.  In a time interval dt a small amount dm of the moisture receives an impulse vdm.  In the same time interval the velocity of the drop changes by dv.  The total momentum change of the system is dp = mdv + vdm.
How do they apply?
Choose a coordinate system such that the z-axis points downward and F = mgk.  The problem then is a one-dimensional problem.
p = mv,  dp/dt = mdv/dt + vdm/dt = mg.  dv/dt = g - (v/m)dm/dt.
m = r4pR3/3,  dm/dt = l4pR2, since the increase in mass per unit time is proportional to the surface area.
But we also have dm/dt = r4pR2dR/dt.  Therefore dR/dt = l/r, R = C + (l/r)t.
Details of the calculation:
dv/dt = (dv/dR)(dR/dt) = (l/r)(dv/dR).
But we also have
dv/dt = g - (v/m)(dm/dt) = g - 3vl/(rR).
Therefore
dv/dR = (rg/l) - 3v/R.
[The differential equation dx/dt + nx/t = B has the general solution
x = At-n + Bt/(n+1).]
v = AR-3 + (rg/4l)R, v = A/(C + (l/r)t)3 + (rg/(4l))(C + (l/r)t).
At t = 0 v = v0.  v0 = A/C3 + rgC/(4l),  A = v0C3- rgC4/(4l).
As C goes to zero A goes to zero, and v =  (rg/(4l))(l/r)t = gt/4.
v increases linearly with time.