Question

A bullet of mass m = 2.40×10-2 kg is fired along an incline and embeds itself quickly into a block of wood of mass M = 1.35 kg. The block and bullet then slide up the incline, assumed frictionless, and rise to a height H = 1.35 m before stopping. Calculate the speed of the bullet just before it hits the wood.

Answer 1

Using the energy conservation after the bullet hits the block and when it reaches at top

KEi + PEi = KEf + PEf

KEi = 0.5*M*V^2

PEi = 0, since h = 0

KEf = 0, since at max height velocity will be zero.

PEf = m*g*H

0.5*M*V^2 + 0 = 0 + M*g*H

V = sqrt (2*g*H)

V = sqrt (2*9.81*1.35) = 5.146 m/sec

Now using momentum conservation before and after the collision

Pi = Pf

m1v1 + m2v2 = MV

m1 = 2.4*10^-2 kg

m2 = 1.35 kg

v1 = ?

v2 = 0 m/sec

M = m1 + m2 = 1.35 + 0.024 = 1.374 kg

v1 = 1.374*5.146/(2.4*10^-2)

v1 = 294.61 m/sec