A bullet with mass m = 5.21 g is moving horizontally with a speed v = 548 m/s wh

Question

                     A bullet with mass m = 5.21 g is moving horizontally with a speed v = 548 m/s when it strikes a block of wood with mass M = 14.8 kg (initially at                     rest). The bullet imbeds itself in the block.                 

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

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

                    (c) What is the final 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?

Explanation / Answer

A) .328 m/s

B) 4.85 kg m/s

C)K.E = 0.5*m*v^2 = 0.5*14.8*(0.328^2) = 0.796 JD) 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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