A bullet with a mass of 4.0 g and a speed of 691 m/s is fired at a block of wood

Question

A bullet with a mass of 4.0 g and a speed of 691 m/s is fired at a block of wood with a mass of 0.099 kg. The block rests on a frictionless surface, and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is 25 m/s. What is the speed of the bullet when it exits the block? Is the final kinetic energy of this system equal to, less than, or greater than the initial kinetic energy? Verify your answer to part (b) by calculating the initial and final kinetic energies of the system.

Explanation / Answer

Intial momentum of the system = momentum of the bullet, P1 = mv1 = 0.004*691 = 2.764 kg.m/s

Now, final momentum of the system = momentum of the bullet + momentum of the block = 0.004*v2 + 0.099*25

(a) Now applying conservation of momentum -

2.764 = 0.004*v2 + 2.475

=> v2 = 72.25 m/s

So, speed of the bullet when it exits the block = 72.25 m/s.

(b) Initial kinetic energy of the system, E1 = 0.5*0.004*691^2 = 954.962 J

Final kinetic energy of the system =  0.5*0.004*72.25^2 +  0.5*0.099*25^2 = 10.44 + 30.94 = 41.38 J

So, final kinetic energy is less than the initial kinetic energy. Last option is correct.

This is due to the fact that some energy is converted into the heat.

(c) As calculated in part (b)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.