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:

Concepts: Inelastic collision, conservation of momentum | |

Reasoning: 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. | |

Details of the calculation: The initial momentum of player 1 is p_{1
}= (90 kg 5 m/s)i = (450 kgm/s)i.The initial momentum of player 2 is p_{2 }= (95 kg 3 m/s)j = (285 kgm/s)j.Momentum is conserved, the final momentum p of both players is p
= p_{1 }+ p_{2}.p = (m_{1}
+ m_{2})v.v = (2.432 m/s)i + (1.54 m/s)j.v ^{2 }= 8.29(m/s)^{2}, v = 2.88 m/s.The speed of the players after the collision is 2.88 m/s. tanθ = p _{y}/p_{x
}= 285/450 = 0.63, θ = 32.34^{o}.Their direction of travel makes an angle θ= 32.34 ^{o} 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:

Concepts: Newton's 2nd law, F = dp/dt | |

Reasoning:F = dp/dt, yields the equations of motion,
which can be solved for the velocity as a function of position. | |

Details of the calculation: (a) F = dp/dt. (1-dimensional problem) p = mv = (M + m _{rope})v, dp = vdm_{rope} + mdv. dm_{rope}
= rdx, m = M + rx.F = rv ^{2} + (M + rx)dv/dt.We are interested in dv/dx. dv/dt = (dv/dx)v. vdv/dx = (F - rv ^{2})/ (M
+ rx).Let x' = (M + rx), then dx' = rdx. vdv/dx' = (F/r– v ^{2})/x'.
Let F/r = B.vdv/dx' = (B - v ^{2})/x', dv^{2}/dx' = 2B/x'
- 2v^{2}/x' = 2(B – v^{2})/x'.Let v' ^{2} = -v^{2}.
Then dv'^{2}/dx' = -2v'^{2}/x'.v' ^{2} = C_{1}/x'^{2}.
dv'^{2}/dx' = -2C_{1}/x'^{3} = -2v'^{2}/x'.
v^{2} = F/r - C_{1}/x'^{2}.The initial conditions are v ^{2} = 0 and x' = M.Therefore C _{1} = M^{2}
F/r.v ^{2}(x) = (F/r)(1 - M^{2}/(M + rx)^{2}).(b) As x--> infinity, v ^{2} --> 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:

Concepts: Elastic collisions, energy and momentum conservation | |

Reasoning: In elastic collisions mechanical energy and momentum are conserved. | |

Details of the calculation: For a collision between a sphere of mass m _{1} and initial velocity
v_{1i }and a stationary sphere of mass m_{2}, energy and
momentum conservation yield v_{2f }= 2m_{1}v_{1i}/(m_{1}
+ m_{2}).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: v _{2f }= 8mv/(4m +
M).Now v _{2f} becomes v_{1i} for the collision between M and m.We have for this collision: v _{f }= 16mMv/[(m + M)(4m + M)], where v_{f}
is the speed of mass m immediately after the collision.We want to maximize v _{f} with respect to M.v _{f} = 16mMv/(M^{2} + 5Mm + 4m^{2}),dv _{f}/dM = 0 --> 16m(M^{2} + 5Mm + 4m^{2}) -
16mM(2M + 5m) = 0.16M ^{2} = 64m^{2}, M = 2m. This is the only physical value for M for which dv _{f}/dM = 0.v _{f }--> 0 as M --> 0 or infinity, we have a maximum.v _{fmax} = (16/9)v = (16/9)(2gh)^{1/2}.H = v _{fmax}^{2}/2g = (16/9)^{2}h = 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 μ_{s} and μ_{k},
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:

Concepts: Newton's second law, friction | |

Reasoning: There are two scenarios. (i) If θ < θ _{c}, where θ_{c} 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 < θ _{c}, 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.) | |

Details 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 θ _{c} is reached when the ribbon tension becomes equal to
the smaller block's static friction limit. In that limiting case we have2g sin θ _{c} [Mm/(M + m)] = μ_{s}mg cosθ_{c}, tan θ_{c}
= (μ_{s}/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 = μ _{k}mg cosθ.The bigger block and ribbon then have downhill acceleration given by a = [Mg sinθ – μ _{k}mg cosθ]/M = g[sinθ – μ_{k}(m/M) cosθ].As the ribbon slides uphill under it, the smaller block has downhill acceleration given by: a = [mg sin θ – μ _{k}mg cos θ]/m = g (sinθ – μ_{k} cos θ).(If μ _{k} > tan θ, then the smaller block's acceleration will be
negative, i.e. uphill.) |

**Problem 5:**

A rocket of initial mass m_{0} ejects fuel at a constant rate km_{0}
and at a velocity v' relative to the rocket shell of mass m_{1}.

(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:

Concept: Motion in one dimension, Newton's 2nd law, F = dp/dt, dp = p(t + dt) - p(t) | |

Reasoning: 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. | |

Details of the calculation: (a) -mg = m(dv/dt) - |dm/dt|v'.Here |dm/dt| = km _{0}. Therefore dv/dt = -g + km_{0}v'/m.At t = 0 m = m _{0}. 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 > m _{1} we
have m = m_{0} - km_{0}t. Thereforedv/dt = -g + km _{0}v'/(m_{0}
- km_{0}t) = -g + kv'/(1 - kt).At time t _{c} the fuel is
used up. m _{1} = m_{0} - km_{0}t_{c}, t_{c}
= (1 - m_{1}/m_{0})/k = (1 - m_{1}/m_{0})v'/g
under the conditions stated in the problem.For t < t _{c}, dv/dt
= -g + kv'/(1 - kt).Let t' = 1 - kt. Then dv/dt' = g/k - v'/t'. We can integrate to find v(t'). ∫ _{0}^{v(t1')}dv = ∫_{t0'}^{t1'}(g/k)dt' - ∫_{t0'}^{t1'}(v/t')dt'
= (g/k)(t_{1}' - t_{0}') - v'ln(t_{1}'/t_{0}').t _{1}'
= 1 - kt_{1}, t_{0}' = 1. v(t_{1}) = -gt_{1}
- v'ln(1 - kt_{1}) = -gt_{1} - v'ln(1 - gt_{1}/v').For t < t _{c}, v(t) = -gt_{ } - v'ln(1 - gt/v').At t _{1} = t_{c}, v(t_{c}) = -v'(1 - m_{1}/m_{0})
- v' ln(m_{1}/m_{0}). This is the greatest speed of the
rocket.At time t _{c} the height of the rocket ish(t _{c}) = ∫_{0}^{tc}v(t)
dt = -gt_{c}^{2}/2 + (v'^{2}/g)∫_{1'}^{1-gtc/v'}ln(t')dt'
= -gt_{c}^{2}/2 + (v'^{2}/g)(t'ln(t') - t')|_{1}^{m1/m0}.h(t _{c}) = (v'^{2}/g)[(m_{1}/m_{0})ln(m_{1}/m_{0})
- ½(m_{1}^{2}/m_{0}^{2}) + ½].For t > t _{c} the rocket is coasting in a gravitational field. It
will climb until its speed has decreased to zero.Δh = ½v(t _{c})^{2}/g.
The total height reached is h_{total} = h(t_{c}) + Δh. |