Practice portal › Laws of Motion › Conservation of Linear Momentum
Asked in NEET 2005 · Recoil and explosions
Given: a 30 kg bomb at rest splits into 18 kg moving at 6 m s⁻¹ and 12 kg.
Idea: the total momentum stays zero, which fixes the speed of the second piece.
0=18×6+12× v, so 12v=-108 and v=-9 m s⁻¹.
The minus sign only says the pieces fly apart in opposite directions.
K=1/2(12)(9)²=1/2×12×81.
=486 J.
Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.
Practise this with a timer