A wooden block of mass M = 5.99 kg hangs from two long massless cords (a ballistic pendulum). A bullet of mass m = 10 g is fired horizontally into the block and embeds itself. The block plus bullet then swing up so that their centre of mass rises h = 9.8 cm before momentarily stopping. Tak

Question

A wooden block of mass M = 5.99 kg hangs from two long massless cords (a ballistic pendulum). A bullet of mass m = 10 g is fired horizontally into the block and embeds itself. The block plus bullet then swing up so that their centre of mass rises h = 9.8 cm before momentarily stopping. Taking g = 9.8 m/s², what was the bullet's speed just before impact?

Accepted Answer

831.4 m/s. The post-collision speed is V = sqrt(2gh). Momentum conservation gives bullet speed u = ((M+m)/m)*sqrt(2gh).

Solution

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