A bullet with mass m = 5.21 g is moving horizontally with a speed v = 466 m/s wh

Question

A bullet with mass m = 5.21 g is moving horizontally with a speed v = 466 m/s when it strikes a block of hardened steel with mass M = 14.8 kg (initially at rest). The bullet bounces off the block in a perfectly elastic collision.

(a) What is the speed (m/s) of the block immediately after the collision?

(b) What is the impulse (kg m/s) exerted on the block?

(c) What is the nal kinetic energy (J) of the block?

(d) How much work (J) did the bullet do on the block?

(e) What was the change in kinetic energy (J) of the bullet?

(f) How much work (J) was done on the bullet?

I know what the answers are and I know how to do (A) and (B). I would like to see the work on how to solve for (C), (D), (E) and (F).

Answers:

A) .328 m/s

B) 4.85 kg m/s

C) 0.796 J

D) 0.796 J

E) -0.796 J

F) -0.796 J

Explanation / Answer

C) K.E = 0.5*m*v^2 = 0.5*14.8*(0.328^2) = 0.796 J D) box was initially at rest so the work done by bullet is equal to change in K.E of the block W = 0.796 J E) since the collision is perfectly elastic no energy loss takes place change in the kinetic energy of bullet = -work done by bullet on the block = -0.796 J F) Work done by block = -change in the kinetic energy of block = -0.796 J

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