A rugby player runs with the ball directly towards his opponent's goal, along the positive direction of an x axis. He can legally pass the ball to a teammate as long as the ball's velocity relative to the field does not have a positive x component. Suppose the player runs at speed 4.0 m/s relative to the field while he passes the ball with a speed of 6.0 m/s relative to himself. What is the smallest angle (relative to the x axis) the ball can be passed in (as seen from the player) in order for the pass to be legal?

Solution:

- Concepts:

Kinematics, frame transformations - Reasoning:

Let the velocity of the player be**v**_{1}and let the relative velocity between the player and the ball be**v**_{2}.

Then the velocity of the ball relative to the field is**v**=**v**_{1}+**v**_{2}=**i**(v_{1}+ v_{2}cosθ) +**j**v_{2}sinθ. - Details of the calculation:

For the smallest angle θ_{min}*,***v**has no x-component, v_{1}+ v_{2}cosθ_{min}= 0.

cosθ_{min}= -v_{1}/v_{2}= -4/6. θ_{min}= 131.8^{o}.

A 10 g bullet is stopped in a block of wood (m = 5 kg). The speed of the bullet-wood combination immediately after the collision is 0.6 m/s. What was the original speed of the bullet?

Solution:

- Concepts:

Inelastic collisions, momentum conservation - Reasoning:

The particle has an inelastic collision with the block. The two objects stick together after the collision and move with a common velocity**v**_{f}. - Details of the calculation:

Assume the motion is along the x-direction. We then have from momentum conservation

m_{1}v_{1 }+ m_{2}v_{2 }= (m_{1}+ m_{2})v_{f},

where m_{1}is the mass of the bullet, v_{1}its initial velocity, m_{2}is the mass of the block, v_{2}its initial velocity. Initially the block is at rest, v_{2 }= 0. Therefore

(0.01 kg)v_{1 }= (5.01 kg) 0.6 m/s. v_{1}= 300.6 m/s.

An object is sliding on a frictionless surface with velocity V_{0}.
There is a track on the surface, which turns object backwards as shown by
the dashed
line in the figure. If the radius of the track is R and the coefficient of
kinetic friction
between object and track is μ, how long will it take for the object to make a 180
degree turn?

The size of the object is negligible compared to the radius of the track.

Solution:

- Concepts:

Circular motion, Newton's 2nd law, friction - Reasoning:

For the particle of mass m to move with speed v on the inside of the circular track, the track has to exert a force mv^{2}/R towards the center of the circle. This is a normal force.

The frictional force with magnitude f = μN = μ mv^{2}/R will slow the particle down. - Details of the calculation:

(a) mdv/dt = -μmv^{2}/R, dv/v^{2}= -(μ/R)dt.

Integrate: ∫_{v0}^{v(t)}dv'/v'^{2}= -(μ/R)t. 1/v_{0}- 1/v(t) = -(μ/R)t. v(t) = v_{0}/(1 + (μv_{0}/R)t).

The distance s(t) traveled in time t is s(t) = ∫_{0}^{t}v(t')dt' = v_{0}∫_{0}^{t}dt'/(1 + (μv_{0}/R)t').

s(t) = (R/μ) ∫_{1}^{1 + (μv0/R)t}dt/t' = (R/μ)ln(1 + (μv_{0}/R)t).

For s(t) = πR we have πμ = ln(1 + (μv_{0}/R)t), exp(πμ) = 1 + (μv_{0}/R)t.

t = R(exp(πμ) - 1)/(μv_{0}).

Or

dv/dt = (dv/ds) ds/dt = ½dv^{2}/ds = - μv^{2}/R.

v^{2}(s) = v_{0}^{2}exp(-2μs/R), v(s) = v_{0}exp(-μs/R), v_{after 1/2 revolution }= v_{0}exp(-πμ).

v(s) = ds/dt dt = ds/v(s) = ds exp(μs/R)/v_{0}.

Integrate from 0 to πR to find t(πR) = (exp(πμ) - 1)R/(μv_{0}).

A small massive bead can slide without friction along a wire ring of radius R mounted in the vertical plane. A light elastic cord is attached to the top of the ring and to the bead.
The relaxed length of the cord is L. If the cord is stretched by ΔL > 0,
it exerts an elastic force F = -kΔL directed toward the attachment point on top
of the ring.

Initially, the bead is at rest at the bottom of the ring, and the
magnitude of the force exerted on the bead by the ring is twice the bead's weight. After a small disturbance, the bead begins to slide upward along the ring, reaching its maximum velocity at the moment it covers one-third of the ring's circumference.

(a) What is the length of the relaxed cord?

(b) What is v_{max}?

(c) Find the force exerted by the ring on the bead when v = v_{max}.

Solution:

- Concepts:

Newton's second law, conservation of energy - Reasoning:

We draw a free-body diagram and find the tangential acceleration and the centripetal acceleration as a function of angle from Newton's second law.

v = v_{max}when a_{t}= 0.

We find v_{max}using energy conservation. - Details of the calculation:

(a) Initially the bead is at rest. The net force is zero, N_{0}+ mg = 3mg = kΔL_{0}. ΔL_{0}= 2R - L is the initial stretch of the cord and N_{0}= 2mg points radially away from the center of the ring.

Now consider an instant in time when the cord makes an angle θ relative to the vertical, as in the diagram. The bead's angular position is labeled φ in the diagram.,

F**N**, and m**g**act on the bead.

We have ma_{t}= F sinθ - mg sinφ, mv^{2}/R = F cosθ - N - mg cosφ.

From geometry we have

180^{o}- φ = 180^{o}- 2θ, θ = φ/2. Rcosθ = (L + ΔL)/2.

So ma_{t}= kΔL sin(φ/2) - mg sinφ, ma_{t}= k(2Rcos(φ/2) - L)sin(φ/2) - mg sinφ.

Given: When φ = 120^{o}, v = v_{max}, a_{t}= 0.

0 = k(2Rcos(60^{o}) - L)sin(60^{o}) - mg sin(120^{o}).

k(L - R) = -mg, L = R - mg/k.

From the initial condition we have L = 2R - 3mg/k.

These two equations together yield L = R/2.

(b) We can find v_{max}from energy conservation. We have to consider gravitational and elastic potential energy.

½mv_{max}^{2}= -mgy_{φ=0}+ mgy_{φ=120}+ ½kΔL^{2}_{φ=0}- ½kΔL^{2}_{φ=120}.

At φ = 0^{o}we have L + ΔL = 2R = R, ΔL = 3R/2, y = 2R.

At φ = 120^{o}we have L + ΔL = 2Rcos(60^{o}) = R, ΔL = R/2, y = Rcos(60^{o}) = R/2.

We also have mg/k = R/2.

So ½mv_{max}^{2}= -3mgR/2 + ½k(3R/2)^{2}- ½k(R/2)^{2}= -3mgR/2 + ½k(2R^{2}).

v_{max}^{2}= gR.

(c) mv^{2}/R = Fcosθ - N - mg cosφ.

At φ = 120^{o}we have mg = kR/4 - N + mg/2.

N = mg/2 + mg/2 - mg = 0.

The density of aluminum is 2.7 g/cm^{3}. A 10 cm diameter
aluminum cable is used to lower a 1000 kg mass into a 300 m deep bore hole.

(a) When the mass is held just above the bottom of the hole, what is the
tension in the cable as a function of depth? At what depth is the tension
largest?

(b) If the yield strength (force per area) of the cable is 10 MPa, what
minimum diameter of the cable could be used?

Solution:

- Concepts:

Tension and equilibrium - Reasoning:

The tension in the cable at depth x must be equal in magnitude to the weight suspended below depth x. -
Details of the calculation:

(a) T(x) = ((300 m - x)*(2700 kg/m^{3})*π*(0.05 m)^{2}+ 1000 kg)*(9.8 m/s^{2}).

T(x) = (27145 - 207.8x )N, with x measured in meter.

T_{max }= 27145 N at depth x = 0.

(b) 10^{7}N/m^{2}*π*r^{2}= T_{max }= ((300 m)*(2700 kg/m^{3})*π*r^{2}+ 1000 kg)*(9.8 m/s^{2}).

π*r^{2}(10^{7}- 300*2700*9.8) N/m^{2}= 9800 N.

r = 0.039 m.

d_{min }= 2r = 7.8 cm is the minimum diameter of the cable that could be used.