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  • A bullet of mass m moving with a velocity v strikes a suspended wooden block of mass M as shown in the figure and sticks to it. If the block rises to a height h then  the initial velocity of t

A bullet of mass m moving with a velocity v strikes a suspended wooden block of mass M as shown in the figure and sticks to it. If the block rises to a height h then  the initial velocity of the bullet is 

Option: 1

\frac{m}{m- M} \sqrt{2gh}


Option: 2

\frac{m}{m + M} \sqrt{2gh}


Option: 3

\frac{m + M}{m} \sqrt{2gh}


Option: 4

\frac{m - M}{m} \sqrt{2gh}


Answers (1)

Collision Between Bullet and Vertically Suspended Block -

A block of mass M suspended by vertical thread.

A bullet of mass m is fired horizontally with velocity u in the block.

                                                               

Let after the collision bullet gets embedded in block.

And, the combined system raised upto height h where the string makes an angle \theta  with the vertical. 

  1. Common velocity of system just after the collision (V)

Here, system is (block + bullet) 

                                     \\ {\text { P=momentum }} \\ \\ {P_{\text {bullet}}+P_{\text {block}}=P_{\text {system}}} \\ \\ {m u+0=(m+M) V} \\\\ {V=\frac{m u}{m+M}}                         .....(1)

  1. Initial velocity of the bullet in terms of h

By the conservation of mechanical energy

\\ {(T . E \text { of system }) \text { Just after collision }=(T . E \text { of system) At height } h} \\\\ {\frac{1}{2}(m+M) V^{2}=(m+M) g h} \\ \\ {V=\sqrt{2 g h}} \\ \\ {\text { Equating }(1) \text { and }(2)} \\ \\ {\text { We get } V=\sqrt{2 g h}=\frac{m u}{m+M}}

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

So, the answer is - 

\frac{m + M}{m} \sqrt{2gh}

Posted by

Sumit Saini

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