On this page
A bullet hits a wooden block hanging on a string, stays inside it, and the block swings up a few centimetres. From that small rise you can work out the bullet's speed. The problem is a favourite because it tests whether you know which conservation law holds at each moment. Use one law for the whole story and you will be wrong by a large factor. This post gives the two-stage method, a full worked example, the variants examiners like and a practice routine.
What happens in a ballistic pendulum?
There are two separate events.
The impact. It lasts a tiny time. The string is vertical and gravity is vertical, so there is no horizontal external impulse, and horizontal momentum is conserved. The NCERT Class 11 chapter Work, Energy and Power states that "in all collisions the total linear momentum is conserved", and calls a collision in which the bodies move together afterwards "completely inelastic". Kinetic energy is not conserved: it turns into heat, sound and deformation of the wood.
The swing. Now gravity and tension act over a long time, so momentum changes. Tension does no work because it is always perpendicular to the motion, so mechanical energy is conserved.
Both ideas are in Unit 4 of the JEE (Main) 2026 syllabus, which lists conservation of mechanical energy and "elastic and inelastic collisions in one and two dimensions". Check your exam year's syllabus.
Worked example: finding the bullet's speed
A 10 g bullet is fired horizontally into a 1.99 kg block hanging from a 1.0 m string and stays embedded. The block rises 0.20 m. Take g = 10 m/s².
Swing (energy). ½(m + M)V² = (m + M)gh, so V = √(2gh) = √(2 × 10 × 0.20) = 2.0 m/s.
Impact (momentum). mv = (m + M)V, so v = (2.00 × 2.0)/0.010 = 400 m/s.
Energy check. Before: ½ × 0.010 × 400² = 800 J. After: ½ × 2.00 × 2.0² = 4 J. NCERT's formula for the loss, ½ × mM/(m + M) × v², gives ½ × (0.010 × 1.99/2.00) × 160,000 = 796 J. Only m/(m + M), or 0.5%, of the bullet's energy survives the impact.
String tension just after impact. T = (m + M)(g + V²/L) = 2.00 × (10 + 4.0) = 28 N, compared with 20 N while hanging at rest.
Exams sometimes give the sideways swing instead of the rise. If the block swings 0.60 m sideways on the 1.0 m string, then h = L − √(L² − x²) = 1.0 − 0.80 = 0.20 m, the same as above.
Why can't you use energy conservation straight through?
Try it: ½mv² = (m + M)gh gives ½ × 0.010 × v² = 4, so v ≈ 28 m/s. The true answer is 400 m/s, about 14 times larger. The factor is √((m + M)/m) = √200 ≈ 14.1, which shows exactly how much energy the impact throws away.
This mix-up is common. Singh and Rosengrant (2003) built a 25-item test of energy and momentum concepts and found that most introductory students had difficulty interpreting these basic principles and applying them in physical situations.
Which variations do examiners like?
Bullet passes through. A 10 g bullet enters at 400 m/s and leaves at 200 m/s; the block is 2.0 kg. Momentum: 0.010 × 400 = 0.010 × 200 + 2.0 × V, so V = 1.0 m/s and the block rises V²/2g = 0.05 m, or 5 cm.
Full vertical circle. For the block and bullet in the worked example to go all the way round on the string, the speed at the bottom must be at least √(5gL) = √50 ≈ 7.07 m/s. The bullet then needs v ≥ 200 × 7.07 ≈ 1414 m/s. On a light rigid rod the condition is √(4gL), giving about 1265 m/s.
Block on a hinged rod. If the block is replaced by a heavy rod pivoted at the top, linear momentum is no longer conserved because the hinge pushes back. Conserve angular momentum about the hinge instead, as in our post on angular impulse on rigid bodies.
For the restitution side of the story, see collisions and momentum problems.
Which mistakes cost marks?
Equating the bullet's kinetic energy to the final potential energy.
Conserving momentum during the swing. Gravity and tension change it.
Using only M, not m + M, after the impact. With a heavy bullet the difference matters.
Forgetting the centripetal term when asked for tension at the bottom.
A practice routine
For every ballistic problem, write the two stages as two headings before any numbers.
Solve the worked example again with h = 0.45 m. Predict: V = 3.0 m/s, so v = 600 m/s.
Do one pass-through problem and one vertical-circle problem.
Compute the fraction of energy lost each time and check it against m/(m + M).
Two-stage problems are easier to keep straight once you can see each stage happen. The Fermi Simulation Lab has interactive simulations for the ideas that are easier to see than to read.
Before your next mechanics test, rework the 10 g bullet example from memory and check each line against this page.
