A bullet of inertia m traveling at speed v is fired into a wooden block that has

Question

A bullet of inertia m traveling at speed v is fired into a wooden block that has inertia 4m and rests on a level surface. The bullet passes through the block and emerges with speed v/3, taking a negligible amount of the wood with it. The block moves to the right but comes to rest after traveling a distance d.

Part A

What is the magnitude of the frictional force between the block and the surface while the block is moving?

Express your answer in terms of some or all of the variables m, v, and d.

Part B

What is the ratio of the energy dissipated as the bullet passes through the block to the energy dissipated by friction between the surface and the bottom face of the block?

Kcollision/Kfriction = ???

Explanation / Answer

a)Let the velocity of the block just after the bullet has emerged be v'

Conserving momentum,

mv=mv/3+4m*v'

v'=v/6

Conserving energy,

0.5*4m*v'^2=umgd

0.5*4m*v^2/36=u*m*g*d

u=v^2/(18gd)

where g is the acceleration due to gravity

b)Energy lost in collision=0.5mv^2-(0.5*m*(v/3)^2+0.5*4m*(v/6)^2)=0.389mv^2

Energy lost to friction=0.5*4m*(v/3)^2=0.222mv^2

Kcollision/Kfriction=0.389mv^2/0.222mv^2=1.75

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