Ballistic pendulum (height)

Problem

A projectile of mass $m = 10\;\mathrm{g}$ and velocity $v = 300\;\mathrm{m/s}$ embeds itself in a wooden block of mass $M = 1{,}5\;\mathrm{kg}$ hanging from a string (ballistic pendulum). Calculate the maximum height reached by the block (with the projectile inside) after the collision.

Solution

The problem has two phases to be treated separately. Phase 1 — collision (inelastic). The projectile embeds itself in the block: momentum is conserved, not kinetic energy. Immediately after the collision the system (block + projectile) has common velocity $V$. With $m = 0{,}010\;\mathrm{kg}$: $m\,v = (m + M)\,V \;\Rightarrow\; V = \frac{m\,v}{m + M} = \frac{0{,}010\cdot 300}{0{,}010 + 1{,}5}\;\mathrm{m/s} = \frac{3{,}0}{1{,}51}\;\mathrm{m/s} \approx 1{,}99\;\mathrm{m/s}$ Phase 2 — rise (conservation of energy). After the collision the block rises like a pendulum: kinetic energy transforms into gravitational potential energy. The mass cancels out: $\tfrac{1}{2}(m+M)\,V^2 = (m+M)\,g\,h \;\Rightarrow\; h = \frac{V^2}{2g}$ Substituting with $g = 9{,}81\;\mathrm{m/s^2}$: $h = \frac{(1{,}99)^2}{2\cdot 9{,}81}\;\mathrm{m} = \frac{3{,}95}{19{,}62}\;\mathrm{m}$ $\ev{h \approx 0{,}20\;\mathrm{m}}$

Links

Topics: Quantità di moto e urti Concepts: Urto anelastico · Conservazione della quantità di moto · Energia cinetica · Energia potenziale gravitazionale Skills: Conservazione della quantità di moto Objects: Pendolo balistico · Proiettile