A 3.78g bullet moving at 300m/s enters and stops in an initially stationary 3.20

Question

A 3.78g bullet moving at 300m/s enters and stops in an initially stationary 3.20kg wooden block on a horizontal frictionless surface. Part A: What's the speed of the bullet/block combination? Part B: What fraction of the bullet's kinetic energy was lost in this perfectly inelastic collision? Part C: How much work was done in stopping the bullet? Part D: If the bullet penetrated 5.00 cm into the wood, what was the average stopping force? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

(a) Speed of the bullet/block combination immediately after the collision

v = (m1 v1 + m2 v2 ) / (m1 + m2)

m1 = 3.78g; v1 = 300 m/s; m2 = 3.2 kg; v2 = 0;

Substitute and find v

Ans : v = .354 m/s

(b) The combined mass = m = 3.20378 kg
kinetic energy lost=170 J

c)work was done in stopping the bullet=170 J

d)

170=Fd=F*.05

F=3400 N

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