A 10 gram bullet is fired horizontally into, and becomes embedded in, a suspended block of wood whose mass is 0.890 kg. (a) How does the speed of the block with the embedded bullet immediately after the collision compare with the initial speed of the bullet? (b) If the block with the embedded bullet swings upward and its center of mass is raised 0.40 m, what was the initial speed of the bullet? (c) Was the collision elastic? If not, what percentage of the initial kinetic energy was lost?© BrainMass Inc. brainmass.com March 4, 2021, 6:14 pm ad1c9bdddf
Let v be the speed of the bullet and V be the velocity of the block just after the bullet gets in to it.
Since linear momentum is conserved always, we can write,
initial momentum of the bullet = momentum of bullet+block after collision
ie, 10gm * v = (10gm+890gm) * V
Thus v = (10gm+890gm) * V/10gm = 900/10 * V = 90*V
this is the answer to question a.
That is, ...
The solution provides quick, clear steps to finding the out the speeds of a bullet shot into a suspended wooden block in order to determine if the collision is elastic or inelastic using words and simple calculations.