Assignment 2, solutions

Problem 1:

A 90 kg 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:

bulletConcepts:
Inelastic collision, conservation of momentum
bulletReasoning:
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.
bulletDetails of the calculation:
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.
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.
tanθ = py/px = 285/450 = 0.63, θ = 32.34o.
Their direction of travel makes an angle θ= 32.34o with the x-axis.  (The x-axis is pointing east.)

Problem 2:

A ball of mass M on a frictionless horizontal table is pulled by a constant horizontal force F.  A coiled up rope of mass per unit length r is attached which uncoils as the body moves. 
(a)  By taking into account the uncoiling of the rope, derive an expression for the speed of the ball as a function of distance traveled.
(b)  Evaluate the terminal speed of the ball, assuming a very long rope.

Solution:

bulletConcepts:
Newton's 2nd law, F = dp/dt
bulletReasoning:
F = dp/dt, yields the equations of motion, which can be solved for the velocity as a function of position.
bulletDetails of the calculation:
(a)  F  = dp/dt. (1-dimensional problem)
p = mv = (M + mrope)v, dp = vdmrope + mdv.  dmrope = rdx,  m = M + rx.
F = rv2 + (M + rx)dv/dt.
We are interested in dv/dx.
dv/dt = (dv/dx)v.  vdv/dx = (F - rv2)/ (M + rx).
Let x' = (M + rx), then dx' = rdx.
vdv/dx' = (F/r v2)/x'.  Let F/r = B.
vdv/dx' = (B - v2)/x', dv2/dx' = 2B/x' - 2v2/x' = 2(B v2)/x'.
Let v'2 = -v2.  Then dv'2/dx' = -2v'2/x'.
v'2 = C1/x'2.  dv'2/dx' = -2C1/x'3 = -2v'2/x'.  v2 = F/r  - C1/x'2.
The initial conditions are v2 = 0 and x' = M.
Therefore C1 = M2 F/r.
v2(x) = (F/r)(1  - M2/(M + rx)2).
(b)  As x--> infinity, v2 --> F/r.

Problem 3:

Three elastic spheres of equal size are suspended on light strings as shown; the spheres nearly touch each other.  The mass M of the middle sphere is unknown; the masses of the other two spheres are 4m and m.  The sphere of mass 4m is pulled sideways until it is elevated a distance h from its equilibrium position and then released.  What must the mass of the middle sphere be in order for the sphere of mass m to rise to a maximum possible elevation after the first collision with the middle sphere?  What is that maximum elevation H?

 

Solution:

bulletConcepts:
Elastic collisions, energy and momentum conservation
bulletReasoning:
In elastic collisions mechanical energy and momentum are conserved.
bulletDetails of the calculation:
For a collision between a sphere of mass m1 and initial velocity v1i and a stationary sphere of mass m2, energy and momentum conservation yield v2f = 2m1v1i/(m1 + m2).
Let mass 4m move with speed v = (2gh)1/2 just before the first collision.
Here we have for the collision between 4m and M: v2f = 8mv/(4m + M).
Now v2f becomes v1i for the collision between M and m.
We have for this collision: vf = 16mMv/[(m + M)(4m + M)], where vf is the speed of mass m immediately after the collision.
We want to maximize vf with respect to M.
vf = 16mMv/(M2 + 5Mm + 4m2),
dvf/dM = 0 -->  16m(M2 + 5Mm + 4m2) - 16mM(2M + 5m) = 0.
16M2 = 64m2, M = 2m. 
This is the only physical value for M for which dvf/dM = 0.
vf --> 0 as M --> 0 or infinity, we have a maximum.
vfmax = (16/9)v = (16/9)(2gh)1/2.
H = vfmax2/2g = (16/9)2h = 3.16 h.

Problem 4:

A light flat ribbon is placed over the top of a triangular prism as shown in the diagram.   Two blocks are placed on the ribbon.   The coefficients of static and kinetic friction between the ribbon and the blocks are μs and μk, respectively.  There is no friction between the ribbon and the prism.  The angle θ and the masses of the blocks m and M are given.  Assuming that M > m, find the acceleration of the ribbon along the prism after the blocks are simultaneously released.  Consider all possible cases.

Solution:

bulletConcepts:
Newton's second law, friction
bulletReasoning:
There are two scenarios.
(i)  If θ < θc, where θc is some critical angle, then neither block will slide on the ribbon.  The ribbon will pull on the blocks (by means of static friction) with a tension force T.   The smaller block then accelerates uphill and the bigger one accelerates downhill with the same acceleration.
(ii)  If θ < θc, then the smaller block will slide while the ribbon sticks to the larger block. (Since the ribbon is massless, the forces of friction exerted on it by each block must always have equal magnitudes.  Therefore, as a function on θ, the static friction limit is always be reached first by the smaller block, and the larger block never slides along the ribbon.)
bulletDetails of the calculation:
(i)  Assume neither block slides on the ribbon.
a = (T mg sinθ)/m = (Mg sinθ T)/M.
Solving for the unknown tension,
T = 2g sinθ[Mm/(M + m)].  Therefore
a = {[2g sinθ][Mm/(M + m)] mg sinθ}/m = (g sinθ)[M m)/(M + m)].
The critical θc is reached when the ribbon tension becomes equal to the smaller block's static friction limit.  In that limiting case we have
2g sin θc [Mm/(M + m)] = μsmg cosθc,  tan θc = (μs/2)(1 + m/M).

(ii)  When the smaller block is sliding, the ribbon tension is equal to the sliding friction between the smaller block and the ribbon, T = μkmg cosθ.
The bigger block and ribbon then have downhill acceleration given by
a = [Mg sinθ μkmg cosθ]/M = g[sinθ μk(m/M) cosθ].
As the ribbon slides uphill under it, the smaller block has downhill acceleration given by:
a = [mg sin θ μkmg cos θ]/m = g (sinθ μk cos θ).
(If μk > tan θ, then the smaller block's acceleration will be negative, i.e. uphill.)

Problem 5:

A rocket of initial mass m0 ejects fuel at a constant rate km0 and at a velocity v' relative to the rocket shell of mass m1.
(a)  Show that the minimum rate of fuel consumption that will allow the rocket to rise at once is k = g/v' where g is the gravitational acceleration. 
(b)  Find the greatest speed achieved and the greatest height reached under that condition.

Solution:

bulletConcept:
Motion in one dimension, Newton's 2nd law, F = dp/dt, dp = p(t + dt) - p(t)
bulletReasoning:
Choose a coordinate axis with the positive direction upward.  The system consists of the rocket body and the fuel.  The speed of 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 = -mg.
bulletDetails of the calculation:
 (a)  -mg = m(dv/dt) - |dm/dt|v'.
Here |dm/dt| = km0.  Therefore dv/dt = -g + km0v'/m.
At t = 0 m = m0.  We need dv/dt ≥ 0 for the rocket to get off the ground, therefore we need  k ≥ g/v'.
(b)  For t > 0 and m > m1 we have m = m0 - km0t.  Therefore
dv/dt = -g + km0v'/(m0 - km0t) = -g + kv'/(1 - kt).
At time tc the fuel is used up. 
m1 = m0 - km0tc, tc = (1 - m1/m0)/k = (1 - m1/m0)v'/g under the conditions stated in the problem.
For t < tc, dv/dt = -g + kv'/(1 - kt).
Let t' = 1 - kt.  Then dv/dt' = g/k - v'/t'.  We can integrate to find v(t').
0v(t1')dv = ∫t0't1'(g/k)dt' - ∫t0't1'(v/t')dt' = (g/k)(t1' - t0') - v'ln(t1'/t0').
t1' = 1 - kt1, t0' = 1.  v(t1) = -gt1 - v'ln(1 - kt1) = -gt1 - v'ln(1 - gt1/v').
For t < tc, v(t) =  -gt  - v'ln(1 - gt/v').
At t1 = tc, v(tc) = -v'(1 - m1/m0) - v' ln(m1/m0).  This is the greatest speed of the rocket.
At time tc the height of the rocket is
h(tc) = ∫0tcv(t) dt = -gtc2/2 + (v'2/g)∫1'1-gtc/v'ln(t')dt' = -gtc2/2 + (v'2/g)(t'ln(t') - t')|1m1/m0.
h(tc) = (v'2/g)[(m1/m0)ln(m1/m0) - (m12/m02) + ].
For t > tc the rocket is coasting in a gravitational field.  It will climb until its speed has decreased to zero.
Δh = v(tc)2/g.  The total height reached is htotal = h(tc) + Δh.