More Problems

Problem 1:

A pebble is dropped into a water well, and the splash is heard 9 s later.  What is the distance from the rim of the well to the water's surface?   Neglect drag and assume the speed of sound is 340 m/s.

Solution:

• Concepts:
  Kinematics
• Reasoning:
  Let the depth of the well be d.  The time it takes the pebble to fall a distance d (neglecting drag) plus the time it take the sound to travel the distance d is 9 s.
• Details of the calculation:
  t = (2d/g)^½ + d/v_s.  Let x = √d.  t = (2/g)^½x + x²/v_s.
  x² + v_s(2/g)^½x - v_st = 0,  x = -½v_s(2/g)^½ + [v_s²/(2g) + v_st]^½ = 17.85 m^½.
  d = 319 m.

Problem 2:

A car of mass m is slowed down by a drag force F = -kv².  How far will the car travel before its speed is halved

Solution:
Newtonian mechanics
F = mdv/dt = m(dv/dx)(dx/dt) = -kv².
dv/dx  = -(k/m)v.  dv/v = -(k/m)dx.
∫_v0^v0/2dv'/ v = -∫₀^R(k/m)dx,  ln(2) = (k/m)R.
The distance the car will travel before its speed is halved is R = m ln(2)/k.

Problem 3:

A hole is drilled straight through the earth, passing through its center.  The mass of the Earth is M  = 6*10²⁴ kg and its radius is R = 6400 km, and G = 6.67*10^-11 Nm²/kg².
(a)  Find the force on a particle of mass m as function of its distance r from the center. Assume that the density of the Earth is constant.
(b)  Write the differential equation for the motion of the particle.
(c)  If you drop the particle in the hole, what is the period of its motion?   Make a numerical estimate.
(d)  What is this type of motion called?

Solution:

• Concepts:
  Newton's law of gravity, Newton's 2^nd law
• Reasoning:
  Assume a tunnel is bored through the center of the Earth.  The gravitational force on a test particle of mass m in the tunnel, a distance r from the center is in the -r direction and its magnitude is found using "Gauss' law".
  (1/m)∫_{closed area}F∙n dA = 4πGM_inside.
• Details of the calculation:
  (a)  F4πr² = m4πG(4/3)πr³ρ,  F = (4/3)Gmπρr = GmMr/R³.
  (b)  For a particle moving in the tunnel we therefore have F = -kr, k = GmM/R³.  The force on the particle obeys Hooke's law. 
  The differential equation for the motion is d²z/dt² = -(k/m)z, where z is the axis of a coordinate system centered at the center of the earth and pointing along the drilled tunnel.
  (c)  The particle will oscillate with angular frequency ω = (k/m)^½ = (GM/R³)^½. 
  Its period is T = 2π/ω = 5085 s = 85 min.
  (d)  This type of motion is called simple harmonic motion.

Problem 4:

An alpha particle (Z_α = 2, m_α = 4 au) is fired with a kinetic energy 9.50 MeV (when far away) towards a free lead nucleus (Z_Pb = 82, m_Pb = 207 au) that is at rest.  What is the distance of closest approach of the two particles if the alpha particle is fired directly at lead nucleus?

• Concepts:
  Energy and momentum conservation
• Reasoning:
  The electrostatic force is a conservative force.  (We are neglecting radiation.)
• Details of the calculation:
  The electric charge of the alpha particle is q₁ = 2q_e and that of the lead nucleus is q₂ = 82q_e. 
  The alpha particle and the nucleus repel each other.  As the alpha particle moves towards the nucleus, the momentum of the center of mass of the system will stay constant.  At the distance of closest approach, the relative velocity of the two particles will be zero zero.
  When the two particle have the same velocity v_f, we have  m_αv_i =  (m_α + m_Pb)v_f, where v_i is the initial velocity of the alpha particle.
  Energy conservation:  ½m_αv_i² = ½(m_α + m_Pb)v_f² + kZ_αZ_Pbq_e²/d_min.
  We have two equations for two unknowns, v_f and d_min.
  kZ_αZ_Pbq_e²/d_min = ½m_αv_i²(1 - m_α/(m_α + m_Pb)) = T_i(1 - m_α/(m_α + m_Pb)) = T_im_Pb/(m_α + m_Pb).
  d_min = kZ_αZ_Pbq_e²(m_α + m_Pb)/(T_im_Pb)
  = 9*10⁹*2*82*(1.6*10^-19)²(211/207)/(9.5*1.6*10^-13)m = 2.54*10-14 m = 25.4 fm.

Problem 5

A cart of mass M has a pole mounted on it as illustrated in the figure.  Assume the pole mass is negligible.  A ball of mass μ hangs by a massless string, of length R, attached to the pole at point P.
(a)  Suppose the cart (of mass M) and the ball are initially at rest, with the ball hanging in its equilibrium position.  Calculate the minimum velocity that must be imparted to the ball for it to rotate in a circle of radius R in the vertical plane.
(b)  Now suppose the cart and ball have initial velocity V towards the right.  The cart crashes into a stationery cart of mass m and sticks to it.  Find the velocity of the system after a collision.  In this part and the next, neglect friction and assume that M, m >> μ.
(c)  Find the smallest value of the initial cart speed for which the ball can go in circles in the vertical plane following a collision.

Solution:

• Concepts:
  Inelastic collisions, Newton's laws, conservation of momentum, circular motion, frame transformations
• Reasoning:
  Before the collision cart 1 of mass M and mass μ move with uniform velocity v₁ = V.  The impulse of the collision changes the velocity of cart 1, and after the collision cart 1 moves with uniform velocity v₂.  The small mass moves with velocity v = v₁ - v₂ with respect to cart 1 immediately after the collision.  After the collision the rest frame of cart 1 is an inertial frame and we can analyze the problem in that frame.
  Note:  We make use that μ is negligible compared to M and m.  After the collision the center of mass of the system moves with constant velocity.  The position of the CM changes if μ oscillates.  We neglect this.
• Details of the calculation:
  (a)  The forces on the ball are the force of gravity and the tension from the string.  If the initial speed is large enough so that the ball gains a height greater than R, then the centripetal acceleration needed to keep the ball moving on the circle is partly provided by gravity and the tension decreases.  When the tension vanishes then the ball behaves like a free particle in a gravitational field.
  The ball can rotate in a circle if the centripetal acceleration v_top²/R > g.
  Then ½μv² = 2μgR + ½μv_top² > 2μgR + ½μgR = (5/2)μgR.  v_min = (5gR)^½.
  (b)  Conservation of momentum:  Mv₁ = (m + M)v₂,  v₂ = Mv₁/(m + M) = MV/(m + M).
  (c)  In the rest frame of cart 1 immediately after the collision we have a ball of mass μ on a string of length R moving with velocity v.
  v = v₁ - v₂ = mV/(M + m).
  When v > (5gR)^½, the ball can go in circles in the vertical plane following a collision.
  Therefore for V > (M + m)v/m  = (M + m)(5gR)^½/m the ball can go in circles in the vertical plane following a collision.