A bullet with mass m = 5.21 g is moving horizontally with a speed v = 484 m/s wh

Question

A bullet with mass m = 5.21 g is moving horizontally with a speed v = 484 m/s when it strikes a block of wood with mass M = 14.8 kg (initially at rest). The bullet embeds itself in the block. What is the speed (m/s) of the bullet-block combination immediately after the collision? What is the impulse (kg m/s) exerted on the block? What is the final kinetic energy (J) of the block? How much work (J) did the bullet do on the block? What was the change in kinetic energy (J) of the bullet? How much work (J) was done on the bullet?

Explanation / Answer

so for elastic collision velocity of approch=velocity of separtion

so vblock-vbullet=484m/s

and applying conservation of momentum

5.21*10-3*484=5.21*10-3*vbullet+14.8*vblock

using these two eqn we find vblock= 0.34m/s and vbullet= -483.66m/s

so a) 0.34m/s

b) impulse=change in momentum=14.8*0.34=5.032kg m/s

c)kinetic energy=0.5*mv2=0.5*14.8*0.342=0.86J

c) work done by bullet = change in energy of the block=0.86J

d) change in kinetic energy of bullet=0.5*5.21*10-3(483.662-4842)=-0.86J

e) work done on bullet=change in energy of bullet= -0.86J

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