A simple pendulum of length $1 \mathrm{~m}$ has a wooden bob of mass $1 \mathrm{~kg}$. It is struck by a bullet of mass $10^{-2} \mathrm{~kg}$ moving with a speed of $2 \times 10^2 \mathrm{~ms}^{-1}$. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Particle $A$ of mass $m _{1}$ moving with velocity $(\sqrt{3} \hat{i}+\hat{j})\, ms ^{-1}$ collides with another particle $B$ of mass $m _{2}$ which is at rest initially. Let $\overrightarrow{ V }_{1}$ and $\overrightarrow{ V }_{2}$ be the velocities of particles $A$ and $B$ after collision respectively. If $m _{1}=2\, m _{2}$ and after collision $\overrightarrow{ V }_{1}=(\hat{ i }+\sqrt{3} \hat{ j })\, ms ^{-1},$ the angle between $\overrightarrow{ V }_{1}$ and $\overrightarrow{ V }_{2}$ is$......^o$

A $100\, kg$ gun fires a ball of $1\, kg$ horizontally from a cliff of height $500 \,m $. It falls on the ground at a distance of $400 \,m $ from bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity $g = 10\,ms^{-1}$ )

A bomb of $12\, kg$ explodes into two pieces of masses $4\, kg$ and $8\, kg$. The velocity of $8\, kg$ mass is $6\, m/sec$. The kinetic energy of the other mass is .............. $\mathrm{J}$

A $^{238}U$ nucleus decays by emitting an $\alpha$ particle of speed $v\,m{s^{ - 1}}$. The recoil velocity of the residual nucleus is (in $m{s^{ - 1}}$)

A bomb is projected with $200\,m/s$ at an angle $60^o$ with horizontal. At the highest  point, it explodes into three particles of equal masses. One goes vertically upward with  velocity $100\,m/sec$, second particle goes vertically downward with the same velocity as  the first. Then what is the velocity of the third one-