Oblique elastic collisions of two smooth round objects

Eur. J. Phys. 39 (2018) 045002 (7pp)

European Journal of Physics

Oblique elastic collisions of two smooth round objects

Carl E Mungan1,3 and Trevor C Lipscombe2

1 Physics Department, US Naval Academy, Annapolis, MD 21402-1363 United States of America 2 Catholic University of America Press, Washington, DC 20064 United States of America

E-mail: mungan@usna.edu and lipscombe@cua.edu

Received 15 December 2017, revised 21 February 2018 Accepted for publication 1 March 2018 Published 20 April 2018

Abstract The scattering angles and speeds resulting from an off-centre elastic collision of a smooth puck or sphere incident on a second one at rest on a frictionless surface are analysed in terms of the impact parameter and mass ratio. Particular attention is paid to finding simple special cases for the final angles or the relations between them that could be explored in homework problems or laboratory experiments. The level of analysis is accessible to an introductory undergraduate physics course.

Keywords: two-dimensional elastic collision, conservation laws, impact parameter, scattering angles

(Some figures may appear in colour only in the online journal)

1. Introduction

The study of off-centre elastic collisions between two smooth pucks or spheres is a standard topic in the introductory mechanics course [1]. In general there are four unknowns (the final speeds and angles of the two objects in the collision plane) but only three equations (conservation of mechanical energy as well as of the x and y components of linear momentum). Thus one can only find relationships between the final speeds and angles, rather than unique solutions for each of them (although at least one article [2] has incorrectly claimed to have found formulas for the final vector velocities that are valid for oblique collisions) unless an additional quantity such as the impact parameter [3] is specified. Nevertheless by considering

special cases, some of these relationships become powerful and memorable results that are

3 Author to whom any correspondence should be addressed.

0143-0807/18/045002+07$33.00 ? 2018 European Physical Society Printed in the UK

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Eur. J. Phys. 39 (2018) 045002

C E Mungan and T C Lipscombe

Figure 1. Initial velocity vector of magnitude (in red) of a light puck of mass m making an oblique collision with a heavy puck of mass M where the heavy puck is initially at rest. The final velocity vectors of the two pucks are indicated in blue, with magnitude u and angle for the light puck and magnitude V at angle f for the heavy puck. The impact parameter is b and the xy coordinate axes are indicated with the origin centred on the initial position of M.

frequently used in physics. A prominent example is when a puck or sphere strikes an identical puck or sphere at rest; the final velocities of the two bodies are then perpendicular to each other, for reasons deftly explained by Ng [4]. Despite efforts [5], however, similarly simple expressions have not been found when the two objects have unequal masses. The primary goal of the present paper is to derive basic expressions for the two final angles in this more general situation and then to deduce some special cases from them that could be used to develop homework problems or laboratory experiments.

2. Relationships for the two scattering angles

Two round pucks make an off-centre elastic collision on an air table [6]. The sides of the pucks are smooth enough that friction can be neglected and thus there is no mechanism to change the rotational kinetic energies; in particular, if the pucks start out with zero angular speed, they will remain that way after the collision. Let the two pucks have masses m and M where M m. Suppose the heavier puck is initially at rest at the origin, as shown in figure 1. Choose the x axis so that the initial velocity of the lighter puck is parallel to the +x direction with a magnitude of . Choose the y axis so that the y coordinate of the puck of mass m before the collision is positive and equal to the impact parameter4 b. After the collision, the lighter puck has a velocity of magnitude u at angle measured counter-clockwise from the x axis, whereas the heavier puck has a velocity of magnitude V making angle f measured clockwise from the x axis. Given that b > 0, it must be true that 0 < q < 180 because the heavy puck in figure 1 applies an impulse having a +y component to the light puck5. The other angle is restricted to 0 < f < 90 because the incoming light puck in figure 1 applies an impulse having a +x component to the heavy puck.

4 An oblique or off-centre collision is one for which b ? 0. In contrast, a head-on or on-centre collision has b = 0 in which case the two final velocity vectors must be along the x axis and so angles and f are restricted to be either 0? or 180?. 5 Provided m is strictly less than M, the angle can take on any value between 0? and 180? by suitably adjusting the value of b. However, if m is exactly equal to M, then cannot exceed 90? because backscattering cannot occur.

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Eur. J. Phys. 39 (2018) 045002

C E Mungan and T C Lipscombe

Conservation of the x-component of linear momentum implies that

mu = mu cos q + MV cos f

(1)

while conservation of the y-component becomes

mu sin q = MV sin f.

(2)

Finally conservation of kinetic energy leads to

1 mu2 = 1 mu2 + 1 MV 2.

(3)

2

2

2

Solve equation (2) for u to obtain

u

=

MV sin f m sin q

.

(4)

Substitute this expression into equation (1) and solve it for to get

u

=

MV sin (q + m sin q

f) .

(5)

Now substitute both of these results into equation (3) and use the trigonometric identity

sin2 A - sin2 B = sin (A + B) sin (A - B)

(6)

to obtain a compact result relating the two final angles to the mass ratio,

sin (q + 2f) = m sin q.

(7)

M

Using the double-angle formula for sine, this equation can be rearranged into a solution for in the form

tan q

=

sin 2f m/M - cos

2f .

(8)

Other elastic scattering formulae relating the angles in various frames of reference to the

mass ratio are equations (15) to (17) in Pi?a [7], equation (19) with (1) in Ramsey [8], and

equations (3)?(107) and (3)?(110) with (3)?(111) in Goldstein [9]. Given that 0 < q < 180, the right-hand side of equation (7) must be positive and

consequently the left-hand side must be as well, so that 0 < q + 2f < 180. Now on the one hand q + 2f > q because f > 0; but on the other hand sin (q + 2f) sin q according to equation (7) because m/M 1. By examining the graph in figure 2, these facts imply that 90 < q + 2f < 180, in which case sin (q + 2f) = sin (180 - q - 2f) where 0 < 180 - q - 2f < 90. Making that substitution in equation (7), its solution for f

becomes

2f

=

180

-

q

-

sin-1

m M

sin

q,

(9)

where one takes the solution of the inverse sine which returns an angle between 0? and 90?. By analogy to the reasoning of Ng [4], figure 3 shows that the scattering angle f of the

heavy puck is related to the impact parameter b according to

sin f = b ,

(10)

D

where D = R + r (which is the average diameter of the two pucks). For example, f = 0 for a head-on collision, whereas f = 90 for a grazing collision. The simple forms of

equations (7) and (10) result from choosing the heavy puck of mass M to initially be at rest.

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Eur. J. Phys. 39 (2018) 045002

C E Mungan and T C Lipscombe

Figure 2. Graph of the sine function of any angle from 0? to 180?. If sin q has the value indicated by the horizontal red dashed line, then could be either of the two angles 1 and 2 marked at the locations of the vertical red dashed lines. However, q + 2f cannot be an angle to the left of 1 because q + 2f > q; furthermore q + 2f cannot lie between 1 and 2 because sin (q + 2f) sin q; therefore q + 2f must be located in the range between q2 90 and 180?, as indicated in green.

Figure 3. Configuration of the two pucks at the instant of collision for impact parameter b. The light puck of mass m and radius r exerts a normal force N that pushes on the initially stationary heavy puck of mass M and radius R so that its final velocity V makes an angle f in the direction of N. The reaction force to N is the equal and opposite normal force N? that the puck of mass M exerts back on the light puck.

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Eur. J. Phys. 39 (2018) 045002

C E Mungan and T C Lipscombe

Figure 4. Contour plot of equation (11) for the scattering angle of the light puck as a

function of the fractional impact parameter b/D and the mass ratio m/M. The contour

lines are labelled in degrees in increments of 20?. One gets scattering that is exactly backward (q = 180) along the left edge of the graph for an on-centre collision (b = 0) when m < M, and exactly forward (q = 0) along the right edge of the plot for a grazing collision (b = D).

Substituting equation (10) into (8) results in

q

=

tan-1

2b

D

m-

M

1 1

+

b2

D2

2

b2 D2

if

m M

1

-

2

b2 D2

(11)

180

-

2b tan-1 D

1-

1

-

b2 D2

m M

-

2

b2 D2

if

m M

<

1-2

b2 D2

,

where one takes the solution of the inverse tangent which returns an angle between 0? and 90?. Equation (11) is plotted in figure 4. In general, the two scattering angles are independent of the launch speed of the light puck because u and V each scale linearly with .

3. Special cases of these results

Consider m = M as in a billiard-ball collision. In that case, equation (9) becomes the familiar result

2f = 180 - 2q f + q = 90

(12)

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