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.

Moisture condenses at the constant rate λ units of mass per unit time on a falling raindrop. If the drop falls from rest and has initial mass M, find the distance it has fallen in time t. Neglect air resistance.

Solution:

- Concepts:

Newton's 2nd law:**F**= d**p**/dt, d**p**= md**v**+**v**dm - Reasoning:

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**v**dm. In the same time interval the velocity of the drop changes by d**v**. The total momentum change of the system is d**p**= md**v**+**v**dm. - Details of the calculation:

Choose a coordinate system such that the z-axis points downward and**F**= mg**k**. The problem then is a one-dimensional problem.

p = mv, dp/dt = mdv/dt + vdm/dt = mg. dv/dt = g - (v/m)dm/dt.

dm/dt = λ is given.

We have to solve the equation dv/dt = g - (λ/m)v = g - vλ/(m_{0}+λt).

Let t' = m_{0}+λt. Then dv/dt' = (1/λ) dv/dt, dv/dt' + v/t' = g/λ.

Look at the equation dx/dt + nx/t = c. If c = 0, the solution can be found by simple integration. [dx/x = -ndt/t, x = At^{-n}.] If c is not equal to zero, the equation has a solution x/t = constant, x = Bt, with B = c/(n+1).

The most general solution is x = At^{-n}+ ct/(n+1).

For our problem we therefore have v = A/t' + gt'/(2λ). A is determined by the initial conditions.

v(t) = A/(m_{0}+λt) + (m_{0}+λt)g/(2λ), v(0) = 0, A = -gm_{0}^{2}/(2λ).

v(t) = (m_{0}+λt)g/(2λ) - gm_{0}^{2}/(2λ(m_{0}+λt)).

Distance fallen in time t:

d(t) = (1/λ)∫_{m0}^{m0+λt}dt'[A/t' + gt'/(2λ)] = (A/λ)ln(t')|_{m0}^{m0+λt}+ gt'^{2}/(4λ^{2})|_{m0}^{m0+λt}

= -[gm_{0}^{2}/(2λ^{2})] ln((m_{0}+λt)/m_{0}) + [g/(4λ)](2m_{0}t + λt^{2}).

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

(a) The coefficient of friction of the tabletop is μ. What is the
maximum value, α_{c}, for which the chain remains stationary?

(b)
If α is larger than α_{c,} 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:

Newton's 2^{nd}law:**F**= d**p**/dt,

Work-kinetic energy theorem: ∫**F**·d**r**= ΔT. - Reasoning:

This is an effectively one-dimensional problem. The forces acting on the chain are gravity, the normal force, and friction.

∫F·dr = ½mv^{2}. - Details of the calculation:

(a) ρ = m/L = linear mass density.

Assume that the coefficients of static and kinetic friction are the same.

The force of static friction is less or equal to ρ(1 - α)Lgμ, the force of kinetic friction is ρ(1 - α)Lgμ.

When α = α_{c}we have

ρα_{c}Lg - ρ(1 - α_{c})Lgμ = 0.

gravity friction

α_{c}- (1 - α_{c})μ = 0, α_{c}= μ/(μ + 1).

(b) ∫_{αL}^{L}F(x)dx = ½mv_{f}^{2}= ½ρLv_{f}^{2}.

F(x) = ρxg - ρ(L - x)gμ = ρg(1 + μ)x - ρLgμ.

∫_{αL}^{L}F(x)dx = ρg(1 + μ)(1 - α^{2})L^{2}/2 - ρLgμ(1 - α)L.

v_{f}^{2}= g(1 + μ)(1 - α^{2})L = 2Lgμ(1 - α).

A 60 kg man standing fixed to the ground, can throw a m = 150 g baseball
horizontally with a maximum speed of v_{0} = 25 m/s.

(a) How
much mechanical energy does he put into the throw?

(b) Assume the man
stands fixed to a 9.25 kg platform which moves frictionless with respect to the
ground. He has five 150 g baseballs and throws one of the baseballs
horizontally as shown, putting in the same amount of mechanical energy as found
in part (a). The collision of the ball with the platform wall is elastic.
What is his velocity (magnitude and direction) after the throw?

(c)
What is the horizontal velocity component of the ball relative to the man?

(d) Assume the man throws the remaining four balls sequentially, always
putting in the same amount of mechanical energy. What will be his final
velocity?

Solution

- Concepts:

Energy and momentum conservation - Reasoning:

If the man is fixed to the ground, the mechanical energy is equal to the kinetic energy of the baseball. If he is fixed to the platform, the mechanical energy is shared between the ball and the recoiling system.

Since the collision of the ball with the platform wall is elastic, the final velocities are the same as those for a system where the ball is thrown in the opposite direction. - Details of the calculation:

(a) E = ½mv_{0}^{2}= 46.875 J.

(b) E = ½MV^{2}+ ½mv^{2}, energy conservation.

MV = -mv, momentum conservation.

Here m is the mass of the baseball, M is the mass of the recoiling system, v is the velocity of the baseball (chose the coordinate system so v points in the positive x-direction) and V is the velocity of M.

(1D motion, the signs of v and V carry the directional information.)

v = -MV/m, E = ½MV^{2}+ ½M^{2}V^{2}/m. V^{2}= 2Em/(M(M + m)).

V = -(2Em/(M(M + m)))^{½}, the man is moving in the negative x-direction.

Let M_{0}be the mass of the man and the platform and the 5 balls, M_{0}= 70 kg. M = M_{0}- m.

V^{2}= 2*(46.875 J)*0.15/(69.85*70) (m/s)^{2}, V = -5.36 cm/s.

(c) v - V = (M/m + 1) (2Em/(M(M + m)))^{½}= (2E(M + m)/(Mm)))^{½}= 25.02 m/s is the relative velocity of the ball with respect to the man.

(d) Before the nth throw, the platform with the man and the remaining balls is an inertial frame with mass M_{0}- (n - 1)m. After the nth throw M_{n}' = M_{0}- nm is the mass of the recoiling system.

The velocity of the recoiling system in this inertial frame is V' = -(2Em/(M_{n}'(M_{n}'+ m)))^{½}.

After 5 balls have been thrown, the velocity of the man with respect to the ground is

V_{f}= -(2Em)^{½}∑_{n=1}^{5}(M_{n}'(M_{n}'+ m))^{-½}.

M_{n}'(M_{n}'+ m) ≈ M_{0}^{2}- 2M_{0}nm + M_{0}m = M_{0}^{2}(1 - (2n - 1)m/M_{0}).

(M_{n}'(M_{n}'+ m))^{-½}≈ M_{0}^{-1}(1 + (n - ½)m/M_{0}).

V_{f}= -(2Em/M_{0}^{2})^{½}∑_{n=1}^{5}(1 + (n - ½)m/M_{0})

= -(2Em/M_{0}^{2})^{½}[5*(1 - ½m/M_{0}) + (m/M_{0})∑_{n=1}^{5}n]

= -(2Em/M_{0}^{2})^{½}[5*(1 - ½m/M_{0}) + (m/M_{0})*15] = -26.92 cm/s.

A flexible chain of length 3L and mass m hangs over a frictionless pulley so
that the length of the vertical parts of the chain on either side of the pulley
is L. After a slight disturbance, the chain begins to slide to the right.
What is the force exerted by the chain on the pulley at the instant the length
of the vertical part of the chain on the right is 3L/2?

Hint: What is
the acceleration of the CM of the chain?

Solution

- Concepts:

Energy conservation, Newton's laws - Reasoning:

The CM of the chain has vertical and horizontal acceleration components. - Details of the calculation:

Consider the situation where a length L of the chain is still in contact with the pulley.

Let λ = m/(3L) be the mass per unit length of the chain.

If the right end of the chain has moved down a distance z, the potential energy of the chain has changed by

ΔU = -λz gz = -λgz^{2}.

The kinetic energy then is T = ½ λ3Lv^{2}= λgz^{2}.

v^{2}= (2/3)gz^{2}/L. v = z(2g/(3L))^{½}.

Find the linear acceleration of the chain:

dv^{2}/dt = 2vdv/dt = 2(2/3)(g/L)zv.

a = (2/3)(g/L)z.

When z = L/2, a = g/3.

Let us now consider the acceleration of the CM of the chain.

The left side is accelerating upward, the right side is accelerating downward, and the section in contact with the pulley has tangential and centripetal acceleration.

By symmetry, the vertical component of the tangential acceleration of this section is zero, while the vertical component of the centripetal acceleration is

-(1/λL)∫_{0}^{π}λRdθ^{ }sinθ^{ }v^{2}/R = -2v^{2}/L.

The vertical component of the acceleration of the CM therefore is is

a_{y}= [λ(L - z)a - λ(L + z)a - λL(2v^{2}/L)]/(3λL) = -2az/(3L) - 2v^{2}/(3L).

When z = L/2, a_{y}= -g/9 - g/9 = -2g/9.

Only the section in contact with the pulley has tangential acceleration. By symmetry, the vertical component of the tangential acceleration is zero, while the horizontal component is

(1/λL)∫_{0}^{π}λRdθ^{ }a sinθ = 2aR/L = 2a/π.

The horizontal component of the acceleration of the CM is

a_{x}= λL(2a/π)/(3λL) = 2a/(3π).

When z = L/2, a_{x}= 2g/(9π).

The net force on the chain is sum of the force exerted by the pulley and the force exerted by gravity**F**_{pulley}- mg**j**_{ }= 2mg/(9π)**i**- 2mg/9**j**.**F**_{pulley}= 2mg/(9π)**i**+ 7mg/9**j**.

By Newton's third law, the force exerted by the chain on the pulley is

F_{chain}= -**F**_{pulley}= -2mg/(9π)**i**- 7mg/9**j**.

|**F**_{chain}| = ((2mg/(9π))^{2}+ (7mg/9)^{2})^{½}= 0.78 mg.