Question 25: Centre of Mass and Rotational Motion

Question

A bullet of mass $m$ hits a mass $$M$$ and gets embedded in it. If the block rises to a height $$h$$ as a result of this collision, the velocity of the bullet before collision is

Accepted Answer

Let the velocity of the bullet before collision be $$v$$, then according to law of conservation of linear momentum,
$$m v=(m+M) v^{\prime}$$ ..... (i)
As the mass $m$ of the bullet gets embedded in the wall, hence velocity of bullet + target just after collision will be same,
$$\begin{aligned}
& \frac{1}{2}(M+m) v^{\prime 2}=(m+M) g h \
& \Rightarrow \quad v^{\prime}=\sqrt{2 g h} \
\end{aligned}$$
Putting the value of $$v^{\prime}$$ in Eq. (i), we get
$$\Rightarrow \quad v=\sqrt{2 g h} \frac{(m+M)}{m}=\sqrt{2 g h}\left(1+\frac{M}{m}ight)$$

Answer

$$v=\sqrt{2 g h}$$

Answer

$$v=\sqrt{2 g h}\left[1+\left(\frac{m}{M}\right)\right]$$

Answer

$$v=\sqrt{2 g h}\left(1+\sqrt{\frac{M}{m}}\right)$$

Answer

$$v=\sqrt{2 g h}\left[1-\left(\frac{m}{M}\right)\right]$$

Centre of Mass and Rotational Motion

A bullet of mass m hits a mass $M and gets embedded in it. If the block rises to a height h$ as a result of this collision, the velocity of the bullet before collision is

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