IPhO 1967 · Theory problem 1

Bullet passing through a ball resting on a column

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A small ball with mass \(M = 0.2\) kg rests on a vertical column with height \(h = 5\)m. A bullet with mass \(m = 0.01\) kg, moving with velocity \(v_0 = 500\) m/s, passes horizontally through the center of the ball (Fig. 1). The ball reaches the ground at a distance \(s = 20\) m. Where does the bullet reach the ground? What part of the kinetic energy of the bullet was converted into heat when the bullet passed trough the ball? Neglect resistance of the air. Assume that \(g = 10\) m/s².

Figure 1

Fig. 1

Official English text of the problem.

Answers

  • Bullet’s landing point: \(\boxed{100\ \mathrm{m}}\) horizontally from the foot of the column, in the bullet’s original direction of motion—\(80\ \mathrm{m}\) beyond the ball.
  • Energy converted into heat: \(\boxed{Q=1160\ \mathrm{J}}\), corresponding to \(\boxed{92.8\%}\) of the bullet’s initial kinetic energy.

Solution

Bullet’s landing point

Treat the ball and bullet as particles and take the penetration time to be negligible compared with the subsequent fall time. The central, horizontal impact produces horizontal translational motion without rotation. During the brief impact, the external horizontal impulse is negligible, so horizontal momentum is conserved.

After penetration, both objects start falling from height \(h\) with zero vertical velocity. Their common fall time is therefore \[ h=\frac12gt^2, \qquad t=\sqrt{\frac{2h}{g}} =\sqrt{\frac{2(5)}{10}} =1\ \mathrm{s}. \]

With no air resistance, the ball’s horizontal velocity \(V\) remains constant. Its measured range gives \[ V=\frac{s}{t}=\frac{20}{1}=20\ \mathrm{m\,s^{-1}}. \]

Let \(v\) be the bullet’s horizontal velocity after passing through the ball. Conservation of horizontal momentum gives \[ mv_0=mv+MV, \] hence \[ v=v_0-\frac{M}{m}V =500-\frac{0.2}{0.01}(20) =100\ \mathrm{m\,s^{-1}}. \]

The bullet’s horizontal range from the column is thus \[ \boxed{x=vt=(100)(1)=100\ \mathrm{m}}. \]

Energy converted into heat

Kinetic energy is not conserved during penetration. In the problem’s idealized model, the decrease in total kinetic energy is converted into heat. The change in gravitational potential energy during the brief penetration is negligible.

Initially, only the bullet has kinetic energy: \[ K_i=\frac12mv_0^2 =\frac12(0.01)(500)^2 =1250\ \mathrm{J}. \]

Immediately after penetration, the total kinetic energy is \[ K_f=\frac12mv^2+\frac12MV^2 =\frac12(0.01)(100)^2+\frac12(0.2)(20)^2 =50+40=90\ \mathrm{J}. \]

Therefore, \[ \boxed{Q=K_i-K_f=1250-90=1160\ \mathrm{J}}. \]

The required fraction of the bullet’s initial kinetic energy is \[ \boxed{\frac{Q}{K_i}=\frac{1160}{1250}=0.928=92.8\%}. \]

The \(40\ \mathrm{J}\) transferred to the ball’s motion must be excluded from the heat.

The solution was checked against the official solution.