Question

A wooden block of mass 10 gm is dropped from the top of a cliff 100 m high. Simultaneously a bullet of same mass is fired from the foot of the cliff vertically upwards with a velocity of $100m{s}^{-1}$. If the bullet after collision gets embedded in the block, the common velocity of the bullet and the block immediately after collision is $(g=10m{s}^{-2})$

• A. $40m{s}^{-1}$ downward
• B. $40m{s}^{-1}$ upward
• C. $80m{s}^{-1}$ upward
• D. zero

Answer 1

Answer: (B)

$40m{s}^{-1}$ upward

Time after which collision takes place

$t=\frac{h}{u}=\frac{100}{100}=1$sec

Initial velocity of the wooden block

$u_1=gt=10\times 1=10m/s$

Initial velocity of the bullet

$u_2=u-gt=100-10\times 1=90m/s$

$m_1{u}_1-m_2{u}_2=(m_1+m_2)V$

$\therefore V=-40m/s$

Hence the velocity immediately after collision will equal to $40m/s$ in the upward direction .