Collision Between Bullet and Vertically Suspended Block
Category : JEE Main & Advanced
A bullet of mass m is fired horizontally with velocity u in block of mass M suspended by vertical thread.
After the collision bullet gets embedded in block. Let the combined system raised upto height h and the string makes an angle \[\theta \] with the vertical.
(1) Velocity of system
Let \[\upsilon \] be the velocity of the system (block + bullet) just after the collision.
\[\text{Momentu}{{\text{m}}_{\text{bullet}}}\text{+ Momentu}{{\text{m}}_{\text{block}}}\text{= Momentu}{{\text{m}}_{\text{bullet and block system}}}\]
\[mu+0=(m+M)v\]
\[\therefore \] \[v=\frac{mu}{(m+M)}\] ...(i)
(2) Velocity of bullet : Due to energy which remains in the bullet-block system, just after the collision, the system (bullet + block) rises upto height h.
By the conservation of mechanical energy \[\frac{1}{2}(m+M){{v}^{2}}=(m+M)gh\] Þ \[v=\sqrt{2gh}\]
Now substituting this value in the equation (i) we get \[\sqrt{2gh}=\frac{mu}{m+M}\]
\[\therefore \] \[u=\left[ \frac{(m+M)\sqrt{2gh}}{m} \right]\]
(3) Loss in kinetic energy : We know that the formula for loss of kinetic energy in perfectly inelastic collision
\[\Delta K=\frac{1}{2}\frac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}\,{{({{u}_{1}}-{{u}_{2}})}^{2}}\]
(When the bodies are moving in same direction.)
\[\therefore \] \[\Delta K=\frac{1}{2}\frac{mM}{m+M}{{u}^{2}}\] [As \[{{u}_{1}}=u\], \[{{u}_{2}}=0\], \[{{m}_{1}}=m\] and \[{{m}_{2}}=M\]]
(4) Angle of string from the vertical
From the expression of velocity of bullet \[u=\left[ \frac{(m+M)\sqrt{2gh}}{m} \right]\] we can get \[h=\frac{{{u}^{2}}}{2g}{{\left( \frac{m}{m+M} \right)}^{2}}\]
From the figure \[\cos \theta =\frac{L-h}{L}=1-\frac{h}{L}\]\[=1-\frac{{{u}^{2}}}{2gL}{{\left( \frac{m}{m+M} \right)}^{2}}\]
or \[\theta ={{\cos }^{-1}}\left[ 1-\frac{1}{2gL}{{\left( \frac{mu}{m+M} \right)}^{2}} \right]\]
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