Question

A wooden block of mass $10g$ is dropped from the top of a tower $100m$ high. Simultaneously, a bullet of mass $10g$ is fired from the foot of the tower vertically upwards with a velocity of $100m{s}^{-1}$. If the bullet is embedded in it, how high will it rise above the tower before it starts falling?$($Consider $g=10m{s}^{-2})$

• A. 80 m
• B. 85 m
• C. 75 m
• D. 10 m

Answer 1

The correct option is C 75 m

Both of them have to cover 100 m before collision using laws of motion.
$\left(\frac{g{t}^2}{2}\right)+(ut-\frac{g{t}^2}{2})=100$
$ut=100t=100/u=1sec.$
velocity after 1 sec. block$=10m/s,$ bullet $=100-1\times 10=90m/s$
Total momentum $=10\times 90+(-10)\times 10=800gm/s$
After collision,
$80=2\ast 10\ast vv=40m/s$ upwards.
rise after bullet is in block $v^2=u^2+2ass=\frac{v^2}{2a}=\frac{1600}{2\ast 10}=80m$
since it has already fallen 5m from tower it will go 75m further high than tower.