Ballistic Pendulum Calculator
Find the speed of a bullet or projectile from a ballistic pendulum: block and projectile masses plus the rise height or swing angle, with momentum and the kinetic energy lost.
How it works
After impact: V = √(2·g·h) · momentum conservation: m·v = (m + M)·V → v = (m + M)/m · √(2gh) · with a swing angle: h = L·(1 − cos θ) · energy lost = M/(m + M) of the initial kinetic energy
A ballistic pendulum splits the problem into two stages. The projectile embeds in the block in a perfectly inelastic collision, which conserves momentum but not kinetic energy. The block and projectile then swing up together, and mechanical energy is conserved during the swing, so the rise height gives their common speed V = √(2gh). Working back through momentum conservation gives the projectile speed. The method, devised by Benjamin Robins in 1742, is a classic problem in OpenStax University Physics and Halliday & Resnick.
Worked example
A 10 g bullet hits a 2 kg block, which rises 10 cm (g = 9.81 m/s²). Just after impact V = √(2 × 9.81 × 0.10) = 1.401 m/s. Momentum conservation gives v = (0.010 + 2) / 0.010 × 1.401 = 281.54 m/s (1,013.6 km/h). The kinetic energy drops from 396.33 J to 1.972 J — 99.50% is lost to deformation and heat, exactly M/(m + M).
Sources: OpenStax University Physics Vol. 1 — Types of Collisions (ballistic pendulum) · HyperPhysics — Ballistic pendulum
Frequently asked questions
Why can’t I use energy conservation for the whole problem?
The collision is perfectly inelastic: the projectile deforms and heats the block, so almost all its kinetic energy is lost. Only momentum is conserved through the impact; energy conservation applies afterwards, during the swing.
How do I get the rise height from the swing angle?
For a string of length L swinging to angle θ from the vertical, the centre of mass rises h = L(1 − cos θ). Choose “Swing angle and string length” and the calculator does this for you.
Why is so much energy lost?
The fraction kept is m/(m + M). With a light bullet and a heavy block that is tiny — 0.5% for 10 g into 2 kg — so over 99% of the bullet’s energy goes into deformation, heat and sound.
Is the result accurate for real firearms?
It is a good classroom estimate. Air resistance, the block’s rotation and the string’s mass make real measurements differ slightly; ballistic chronographs are used for precise velocities.
Related calculators
Results assume the idealised textbook model stated above; real systems have losses and tolerances, so verify safety-critical values with measurements or a qualified engineer.