JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Conservation of Momentum

A large block of wood of mass

M = 5.99 kg

is hanging form two long massless cords. A bullet of mass

m =

10 g is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their centre of mass rising a vertical distance

h = 9.8 cm

before the (block + bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is :
(take

g = 9.8 ms^{- 2}

)

(A) $821.4 m / s$
(B) $811.4 m / s$
(C) $831.4 m / s$ ✓ Correct
(D) $841.4 m / s$

Answer: (C)

Solution

The bullet embeds itself, so the collision is perfectly inelastic and momentum conservation fixes the common speed

V

of block and bullet immediately afterwards:

m v = (M + m) V

Once they move together no further mechanical energy is lost, so the kinetic energy at the lowest point becomes gravitational potential energy at the highest point:

\frac{1}{2} (M + m) V^{2} = (M + m) g h \Rightarrow V = 2 g h = 2 \times 9.8 \times 0.098 = 1.386 m s^{- 1}

Substituting this back,

v = \frac{M + m}{m} V = \frac{6.00}{0.010} \times 1.386 m s^{- 1}

so the bullet was travelling at about

831.4 m s^{- 1}

immediately before it struck the block.

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Conservation of Momentum →

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Conservation of Momentum

Solution

More JEE Main questions on Conservation of Momentum