Question

A wood block of Mass M=3.00 kg and rests horizontally at the bottom of a ramp. A bullet of mass m=0.0420kg with an intial velocity v0 is fired at the wood block and embedded inside the block. The wood blocl and the embedded bullet together move up the ramp and reach a vertical heigh of yf= 2.40m relative to the bottom of the rmap(yo=0) before sliding downward. Ignore friction/air resistance.

a) What is the total mechanical energy of the block+bullets at the top of the ramp relative to yo=0 b) What is the speed of the block+bullets right after the bullet hits the block c)What is the intial velocity, v0, of the bullet d) How much mechanical energy is lost before and after the bullet hits the block.

Answer 1

let initial velocity = vo

let after collision combined velocity = v

given that The wood blocl and the embedded bullet together move up the ramp and reach a vertical heigh of yf= 2.40m

so energy conservation for this system

KEi + PEi = KEf +PEf

1/2(m+M)v² + 0 = 0 + (m+M)gh

0.5x(0.042+3)v² =  (0.042+3)x9.81x2.4

v = 6.86 m/s .............asswer b

a) total ME = (0.042+3)x9.81x2.4 = 71.62 J

now the collision study b/w block and bullet

applying momentum conservation

m(v_o) + 0 = (m+M)v

v_o = (0.042+3)x6.86 /0.042

v_o= 496.86...........ansc

ans

d lose in ME = (MEf) - (MEi)

= 1/2 m v_o² - 1/2 (m+M) v²  

= 5184.3-71.6

lost ME= 5112.7 J

answer...d