A bullet with a mass m=12.5 g and speed v=84.8 ~m/s is fired into a wooden block

Question

A bullet with a mass m=12.5 g and speed v=84.8 ~m/s is fired into a wooden block with M=113 g which is initially at rest on a horizontal surface. The bullet is embedded into the block. The block-bullet combination slides across the surface for a distance d before stopping due to friction between the block and surface. The coefficients of friction are s=0.753 and k=0.659.

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

m/s ( ± 0.02 m/s)

(b) What is the distance dd?

m ( ± 0.02 m)

Explanation / Answer

Mass of bullet m = 12.5 g

speed of bullet v = 84.8 m/s

Mass of block M = 113 g

speed of block V = 0

a)

From the conservation of momentum, the speed after collision = mv/(m+M) = 12.5*84.8/(12.5+113) = 8.45 m/s

b)

the friction will be kinetic hence the accelereration due to friction force willbe k*g = 0.659*9.81 = 6.46 m/s^2

hence the distance d = v'^2/2a = 8.45^2/(2*6.46) = 5.53 m

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