A 4.4 g bullet is fired into a block of wood with a mass of 21.3 g. The wood block is initially at rest on a 1.3 m tall post. After the collision, the wood block and bullet land 2.4 m from the base of the post. Find the initial speed of the bullet.

Answer 1

m1 =4.4•10^-3 kg, m2 = 21.3•10^-3 kg, h =1.3 m, L= 2.4 m., v1 =?

m1 •v1 = (m1+m2) •v,
v= m1 •v1 /(m1 + m2) .

h =g•t²/2 => t = sqrt(2•h/g)
L= v•t = v• sqrt(2•h/g),
v =L/sqrt(2•h/g),

m1 •v1 /(m1 + m2) = L/sqrt(2•h/g).

v1= L•(m1 + m2)/m1• sqrt(2•h/g) =
= 2.4•(4.4•10^-3+21.3•10^-3)/( 4.4•10^-3)•sqrt(2•1.3/9.8) =
= 2.4•2.57•10^-2/(4.4•10^-3)•0.515 =27.22 m/s.

Answer 2

Elena is right, but MAKE SURE that in the final step when calculating you put the (m1 times sqrt(2 times h/g)) all in parentheses.