A 3 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60^{o}
with the surface. It bounces off with the same speed and angle.
If the ball is in contact with the wall for 0.2 s, what is the average force
exerted on the ball by the wall?

Solution:

- Concepts:

Impulse: d**p**=**F**dt, ∆**p**=**F**_{avg}∆t - Reasoning:

In the elastic collision with the "infinitely" massive wall, the ball receives an impulse ∆**p**. Enough information is given to calculate ∆**p**. Since the contact time is given,**F**_{avg}can be calculated. - Details of the calculation:

The balls initial momentum is**p**_{i }= p_{xi}**i**+ p_{yi}**j**= (3 kg 10 m/s)sin60^{o}**i**+ (3 kg 10 m/s)cos60^{o}**j**.

Its final momentum is

**p**_{f }= p_{xf}**i**+ p_{yf}**j**= -(3 kg 10 m/s)sin60^{o}**i**+ (3 kg 10 m/s)cos60^{o}**j**.

∆**p**=**p**_{f }-**p**_{i}= -2(30 kgm/s)sin60^{o}**i**= -(51.96 kgm/s)**i**.

∆**p**=**F**_{avg}∆t.**F**_{avg }= -(51.96 kgm/s)**i**/(0.2s) = -259.8 N**i**.

A block of mass m_{1} and initial velocity v_{1} collides
head-on with a stationary block of mass m_{2}. The mass m_{2}
compresses a spring of spring constant k. Neglect friction and assume the
collision is elastic.

(a) What is the velocity v_{2}' of m_{2} just after the
collision?

(b) What is the maximum compression of the spring?

Solution:

- Concepts:

Elastic collisions, energy and momentum conservation - Reasoning:

In elastic collisions energy and momentum are conserved. - Details of the calculation:

(a) Let v_{1}'and v_{2}' be the velocities of m_{1}and m_{2}just after the collision.

Momentum conservation: m_{1}v_{1}= m_{1}v_{1}' + m_{2}v_{2}'.

Energy conservation: m_{1}v_{1}^{2}= m_{1}v_{1}'^{2}+ m_{2}v_{2}'^{2}.^{ }v_{2}' = 2v_{1}/(1 + m_{2}/m_{1}).

(b) ½m_{2}v_{2}'^{2}= ½kd^{2}, where d is the maximum compression of the spring

d = (m_{2}/k)^{½}2v_{1}/(1 + m_{2}/m_{1}).

A (non-relativistic) neutron in a reactor makes an elastic head-on collision with the nucleus of
a carbon atom initially at rest.

(a) What fraction of the neutron's kinetic
energy is transferred to the carbon nucleus?

(b) If the initial kinetic
energy of the neutron is 1.6*10^{-13 }J, find its final kinetic energy
and the kinetic energy of the carbon nucleus after the collision.

(The mass
of the carbon nucleus is about 12 times the mass of the neutron.)

Solution:

- Concepts:

Elastic collisions, momentum conservation, energy conservation - Reasoning:

The collision of the two particles is elastic.

Momentum conservation:

(i) m_{1}v_{1i}= m_{1}v_{1f }+ m_{2}v_{2f}.

Energy conservation:

(ii) ½m_{1}v_{1i}^{2}= ½m_{1}v_{1f}^{2 }+ ½m_{2}v_{2f}^{2}. - Details of the calculation:

(a) We can solve this system of two equations for the ratio v_{2f}/v_{1i}. We obtain

v_{2f}^{2 }= (m_{1}/m_{2})(v_{1i}^{2 }- v_{1f}^{2}) from (ii),

v_{1f }= v_{1i }- (m_{2}/m_{1})v_{2f}from (i).

We can therefore write

v_{1f}^{2 }= v_{1i}^{2 }+ (m_{2}/m_{1})^{2}v_{2f}^{2 }- 2(m_{2}/m_{1})v_{1i}v_{2f}.

v_{2f}^{2 }= (m_{1}/m_{2})( 2(m_{2}/m_{1})v_{1i}v_{2f}- (m_{2}/m_{1})^{2}v_{2f}^{2}) = 2v_{1i}v_{2f}- (m_{2}/m_{1})v_{2f}^{2}.

(1 + (m_{2}/m_{1}))v_{2f }= 2v_{1i}.

v_{2f }= 2m_{1}v_{1i}/(m_{1}+m_{2}).

v_{2f}/v_{1i }= 2m_{1}/(m_{1}+m_{2}) = 2/13 = 0.15.

The fraction of the kinetic energy transferred to the carbon nucleus is

(m_{2}/m_{1})(v_{2f}/v_{1i})^{2 }= 12(0.15)^{2 }= 0.284.(b) The initial kinetic energy of the of the neutron is 160 fJ and the final kinetic energy of the carbon nucleus is 45.4 fJ. (1 femtoJoule = 1 fJ = 10

^{-15 }J.) The final kinetic energy of the neutron is (160 - 45.4) fJ = 114.6 fJ.

A non-relativistic neutron with mass m and kinetic energy T scatters elastically
off a nucleus with mass M. What is the maximum kinetic energy that can be
transferred to the nucleus in one collision if

(a) M is initially at rest,

(b) or M is initially allowed to move?

Solution:

- Concepts:

Elastic collisions, energy and momentum conservation - Reasoning:

The maximum kinetic energy is transferred in a head-on collision. We use energy and momentum conservation. - Details of the calculation:

(a) If M is at rest and m = m_{1}, M = m_{2}.

Momentum conservation:

(i) m_{1}v_{1i}= m_{1}v_{1f }+ m_{2}v_{2f}.

Energy conservation:

(ii) ½m_{1}v_{1i}^{2}= ½m_{1}v_{1f}^{2 }+ ½m_{2}v_{2f}^{2}

v_{2f}^{2 }= (m_{1}/m_{2})(v_{1i}^{2 }- v_{1f}^{2}) from (ii),

v_{1f }= v_{1i }- (m_{2}/m_{1})v_{2f}from (i).

We can therefore write

v_{1f}^{2 }= v_{1i}^{2 }+ (m_{2}/m_{1})^{2}v_{2f}^{2 }- 2(m_{2}/m_{1})v_{1i}v_{2f}.

v_{2f}^{2 }= (m_{1}/m_{2})( 2(m_{2}/m_{1})v_{1i}v_{2f}- (m_{2}/m_{1})^{2}v_{2f}^{2}) = 2v_{1i}v_{2f}- (m_{2}/m_{1})v_{2f}^{2}.

(1 + (m_{2}/m_{1}))v_{2f }= 2v_{1i}.

v_{2f }= 2m_{1}v_{1i}/(m_{1}+m_{2}).

The maximum amount of kinetic energy transferred is ½Mv_{2f}^{2}= [4mM/(M+m)^{2}]T.

(b) If M is allowed to move, then the maximum amount of kinetic energy transferred is T, and m is at rest after the collision.

(For this to happen M must initially moves with speed ½(1 - m/M)v into the same direction as m.)

Two perfectly elastic balls, the larger of mass M and the smaller of mass m, with m << M, are dropped from a height h >> radius of either ball above a solid surface. Mass M rebounds elastically from the surface and mass m rebounds elastically from M. Find the height H that the small ball reaches in terms of h. You need only find an approximate result in the limit that m/M --> 0. The general result depends upon the ratio m/M.

Solution:

- Concepts:

Elastic collisions, frame transformations - Reasoning:

When an object of mass m_{1}<< m_{2}and velocity**v**collides head on with an object of mass m_{2}at rest, then m_{1}rebounds with a velocity nearly equal to -**v**. - Details of the calculation:

Assume M collides with velocity**v**= -(2gh)^{½}**j**= -v**j**with the ground. It rebounds with velocity v**j**. In a frame moving with velocity v**j**equal to the velocity of M right after the collision with the ground, m approaches with velocity -2v**j**and rebounds with velocity 2v**j**. In the lab frame m moves with velocity 3v**j**= (2gh')^{½}**j**right after its collision with M, and therefore rises to a height of h' = 9h.

An elastic ball is dropped on a long inclined plane.^{ }It
bounces, hits the plane again, bounces, and so on.^{ }Let us label the
distance between the points of the^{ }first and the second hit d_{12}
and the distance between^{ }the points of the second and the third hit d_{23}.
Find the ratio d_{12}/d_{23}.

Solution:

- Concepts:

Elastic collisions, motion with constant acceleration - Reasoning:

The component of the acceleration perpendicular to the inclined plane and the component of the acceleration parallel to the inclined plane are constant. The balls motion perpendicular to the inclined plane is a series of bounces from some initial height above the plane to the plane and back to the initial height. The time interval between successive contacts with the plane is constant. But the ball also accelerates in the direction parallel to the plane, so the distance along the plane between successive contacts with the plane increases. - Details of the
calculation:

Assume the inclined plane makes an angle θ with the horizontal. Orient the axes of your coordinate system as shown in the figure. The x-axis makes an angle θ with the horizontal and the y-axis is perpendicular to the inclined plane. The ball is dropped vertically and just after the first bounce its velocity vector makes an angle of 90^{o }- θ with the x-axis. Let x = 0, y = 0, t = 0 be the coordinates of the ball at the instant the ball makes contact with the inclined plane the first time.

In the time interval between the first and second bounce the equations giving the position and velocity of the ball as a function of time are:

x(t) = v sinθ t + ½ g sinθ t^{2}, y(t) = v cosθ t - ½ g cosθ t^{2 }v_{x}(t) = v sinθ + g sinθ t, v_{y}(t) = v cosθ - g cosθ t

The ball bounces for the second time when y(t) = 0, t = t_{a}= 2v/g.

x(t_{a}) = (2v^{2}/g) sinθ + (2v^{2}/g) sinθ = (4v^{2}/g) sinθ

v_{x}(t_{a}) = v sinθ + 2v sinθ = 3v sinθ, v_{y}(t_{a}) = v cosθ - 2v cosθ = -v cosθ

Just after the second bounce, the velocity of the ball is v_{x}= 3v sinθ, v_{y}= v cosθ.

Let t' = t – t_{a}. Between the second and the third bounce the equations giving the position and velocity of the ball as a function of time are:

x(t) - x(t_{a}) = 3v sinθ t' + ½ g sinθ t'^{2}, y(t) = v cosθ t' - ½ g cosθ t'^{2 }The ball bounces for the third time when y(t) = 0, t' = t_{b}= 2v/g.

x(t_{a}+ t_{b}) - x(t_{a}) = (6v^{2}/g)sinθ + (2v^{2}/g) sinθ = (8v^{2}/g) sinθ

Therefore d_{12}/d_{23}= x(t_{a})/( x(t_{a}+ t_{b}) - x(t_{a})) = ½.

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.