Question

A hunter wanted to determine the muzzle velocity of a firearm and asked a physics student to help out. They drilled a hole in a table and placed a copy of the physics book (hardcover edition, mass 2.2 kg) over the hole. They fired a 14-gram bullet directly upward through the hole and it became embedded in the book. The book and bullet rose above the table before falling back down. It all happened very fast, but the helper made a video record and they later determined that the maximum height reached by the book and bullet was 29 cm. What was the speed of the bullet when it struck the book?

Answer 1

Using energy conservation after the bullet embedded,

KEi + PEi = KEf + PEf

here, KEi = 0.5*(M + m)*V^2

PEi = 0

KEf = 0

PEf = (M + m)*g*h

given,

M = mass of book = 2.2 kg

m = mass of bullet = 14 gm = 0.014 kg

V = initial speed of book and bullet = ??

g = 9.81 m/s^2

h = rise in height = 29 cm = 0.29 m

So,

0.5*(2.2 + 0.014)*V^2 + 0 = 0 + (2.2 + 0.014)*9.81*0.29

V = sqrt(2*9.81*0.29) = 2.38533 m/s

now using momentum conservation in inelastic collision,

m*u + M*U = (M + m)*V

here, u = initial speed of bullet = ??

U = initial speed of book = 0

then,

0.014*u + 2.2*0 = (2.2 + 0.014)*2.38533

u = 2.214*2.38533/0.014 = 377.22

u = 377 m/s = speed of bullet when it struck the book

"Let me know if you have any query."