m₁ = 50 g = 0.05 kg; m₂ = 450 g = 0.45 kg
The speed of the bullet is u₁. The second body is at rest (u₂ = 0). Let the common velocity of the bullet and the object after the bullet is embedded into the object is v.

\(v=\frac { { m }_{ 1 }{ u }_{ 2 }+{ m }_{ 2 }{ u }_{ 2 } }{ ({ m }_{ 1 }+{ m }_{ 2 }) } \)
\(v=\frac { 0.05{ u }_{ 1 }+(0.45\times 0) }{ (0.05+0.45) } =\frac { 0.05 }{ 0.50 } { u }_{ 1 }\)
The combined velocity is the initial velocity for the vertical upward motion of the combined bullet and the object. From second equation of motion,
\(v=\sqrt { 2gh } \)
\(v=\sqrt { 2\times 10\times 1.8 } =\sqrt { 36 } \)
v = 6 ms^-1
Substituting this in the above equation, the value of u₁ is
\(6=\frac { 0.05 }{ 0.50 } { u }_{ 1 }=\frac { 0.50 }{ 0.05 } \times 6=10\times 6\)
u₁ = 60 ms^-1

Work done, \(W-\int^{x_f}_{x_i}F(x)dx=k\int_0^4x^2 dx={64\over 3}Nm\)

(i) Both charged particles shall be dissimilar charge. (i.e. positive and negative)
(ii) After collision the charged particles should stick together permanent.
(iii) They should move with common velocity.

\(\text {Mass of a bullet } m=20 g=20 \times 10^{-3} \mathrm{~kg} =0.02 \mathrm{~kg}  \)
\(\text {Mass of a pendulum } M =5 \mathrm{~kg}  \)
\(\text {Height } \mathrm{h} =10 \mathrm{~cm}  \)
\(=10 \times 10^{-2}  \)
\(=0.1 \mathrm{~m}  \)

\(\text { K.E. of the block } =\text { P.E. of the block }  \)
\(\frac{1}{2} M v^{2} =M g h  \)
\(\therefore v^{2} =\sqrt{2 g h}  \)
\(=\sqrt{2 \times 9.8 \times 0.1}=\sqrt{1.96}=1.4 \mathrm{~m} / \mathrm{s}  \)
\(\therefore \text { Final speed } v =1.4 \mathrm{~m} / \mathrm{s}  \)
\(\text {Final speed } v =\frac{m_{1} u_{1}+m_{2} u_{2}}{\left(m_{1}+m_{2}\right)}  \)
\(v_{1} =\frac{0.02 u_{1}+5 \times 0}{(0.02+5)}=\frac{0.02}{5.02} u_{1}  \)
\(\text { But } v =1.4  \)
\(\therefore 1.4 =\frac{0.02}{5.02} u_{1}  \)
\(u_{1}=\frac{1.4 \times 5.02}{0.02}=\frac{7.028}{0.02}=351.4 \mathrm{~m} / \mathrm{s}\)
\(\therefore \text { Initial speed } =351.4 \mathrm{~m} / \mathrm{s}\)