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?

Solution Preview

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, ...

Solution Summary

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.

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... Clearly for elastic collision e = 1, for perfectly inelastic collision, the two bodies stick together and moves with the same velocity and hence e = 0. for ...

... That is, total momentum of the bodies before and after collision is same. This is applicable irrespective of whether the collision is elastic or inelastic. ...

... Assuming an elastic collision, determine their two velocities ... paying attention collide in a completely inelastic collision. ... first determine the collisions at x ...

... In inelastic collision the energy is not conserved but ... combined mass just after the collision is given ... energies will decrease and the elastic potential energy ...