Slow collisions in the ballistic pendulum: A computational study

See for example, R. A. Serway and R. J. Beichner, Physics for Scientists and Engineers (Saunders, Philadelphia, 2000), 5th ed., p. 263;

D. C. Giancoli, Physics, Principles with Applications (Prentice–Hall, Upper Saddle River, NJ, 1998), 5th ed., pp. 191–192.

An older reference is Marsh W. White and Kenneth V. Manning, Experimental College Physics (McGraw–Hill, New York, 1954), 3rd ed., pp. 79–86.

We treat the block as a rigid body. This assumption may cause some concern because the bullet penetrates the block. These conditions may be reconciled by imagining that a small-diameter cylindrical channel is drilled through the block, parallel to its long axis and passing through its center of mass. The bullet is assumed to just fit through the channel, and to experience a constant frictional force as it moves.

The structure of the ballistic pendulum analyzed in this paper is similar to what is typically discussed in introductory textbooks (see Ref. 1). However, the standard commercial ballistic pendulum intended for the intro-ductory physics laboratory (for example, as supplied by CENCO or Beck) has a different design: there the pendulum rotates about a fixed axis as a physical pendulum. For this design a somewhat modified analysis is necessary, but it still turns out that the standard treatment does not tell the whole story.

For an interesting discussion, see

T. R.

Sandin

, “,” , – ().

K. R. Symon, Mechanics (Addison–Wesley, Reading, MA, 1971), 3rd ed., p. 189, Eq. (4.127).

This situation could be achieved physically by drilling a cylindrical channel through the block, as described in Ref. 2.

Each supporting rod rotates about a fixed axis (the upper hinge) as the block moves along its circular path. If the rods had mass, additional forces in the direction tangential to the block’s path (perpendicular to the rod) would exist between the block and each rod. Assuming that the rods have zero mass ensures that in this model, the rods exert only tension forces on the block. We also observe that only the sum of the two tension forces enters Eq. (8) for the motion of the center of mass of the block. Although not needed for the present analysis, the individual tension forces of the two rods can be found by applying to the block the relation between angular momentum and torque around the center of mass [see Ref. 4, p. 164, Eq. (4.28)]. Note that the angular momentum of the block around its center of mass is zero for the entire motion.

All numerical values are assumed to be in SI units.

The force that accelerates the block along its circular path is the tangential component $(F cos θ)$ of the frictional force F exerted by the bullet. This force component is smaller for the larger angles θ involved in slow collisions, resulting in smaller values of $V2f.$ The collision time increases for slower collisions, but this effect (which would tend to increase $V2f)$ is relatively small and is outweighed by the reduction in $F cos θ.$

Reference 4, p. 104, Eq. (3.133);

p. 167, Eq. (4.40).

The fact that the tension and normal forces do no work is typical for such reaction forces that arise due to constraints on the motion of a system. See Ref. 4, p. 372.

Referring to Fig. 3, for a slow collision the $V2x$ curve $(d′)$ dips lower than curve d for a fast collision. Hence, the area enclosed between the $V1x$ and $V2x$ curves is larger. Because this area is equal to the magnitude of the frictional work $|WF|$ (see end of Sec. II), it follows that the final kinetic energy of the system is smaller for a slower collision, in agreement with the decrease in $V2f$ previously noted.

Reference 4, p. 161. Eq. (4.7).

Also, the energy and momentum relations work out similarly to those already discussed.

For faster collisions (larger $F),$ the block accelerates through the entire collision, resulting in concave upward behavior for the $θ(t)$ graph until the collision ends, as in curve b of Fig. 5 (for $F=25).$ With slower collisions (smaller $F),$ the gravitational force causes the block to decelerate and the $θ(t)$ graph to turn concave downward even before the collision ends; this effect is visible in curve c (for $F=6).$ Detailed computations show that for this curve, the point of inflection where the concavity changes is at θ=34.6°, while the collision does not end until θ=54.1°. For very slow collisions $(F<4),$ the collision actually ends after the block has reached its highest point and is swinging back.

A bullet fired from a 22 long rifle was found to penetrate 5.7 cm into a pine block. We assume a demonstration apparatus using this bullet $(m1=0.00253$ kg) and a 30 cm×7.5 cm×7.5 cm pine block $(m2=0.843$ kg), with initial bullet velocity $V10=300$ m/s. We keep $R=0.30$ m, as in Table I. The measured penetration depth corresponds to a frictional force $F=1997$ N. By applying our model with these parameters, we find that the collision ends at an angular displacement of only 0.032° (a very fast collision) and the pendulum rises to a maximum angle of 30.318°. The standard treatment gives 30.349° for the maximum angle, which differs by only 0.10% from the prediction of our model.