JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 10)
A block of mass m = 1 kg slides with velocity v = 6 m/s on a frictionless horizontal surface and
collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings
as a result of the collision making angle $$\theta $$ before momentarily coming to rest. If the rod has mass
M = 2 kg, and length $$l$$ = 1 m, the value of $$\theta $$ is approximately :
(take g = 10 m/s2)_3rd_September_Morning_Slot_en_10_1.png)
(take g = 10 m/s2)
63o
69o
55o
49
Explanation
_3rd_September_Morning_Slot_en_10_3.png)
Applying Angular momentum conservation
mv$$l$$ = $$\left( {{{M{l^2}} \over 3} + m{l^2}} \right)$$$$\omega $$
$$ \Rightarrow $$ $$\omega = {{1 \times 6 \times 1} \over {{2 \over 3} + 1}}$$ = $${{18} \over 5}$$
Now, using energy conservation
$${1 \over 2}I{\omega ^2} = \left( {m + M} \right)gh$$
$$ \Rightarrow $$ $${1 \over 2}\left( {{{M{l^2}} \over 3} + m{l^2}} \right){\omega ^2} = \left( {m + M} \right)g{x_{COM}}\left( {1 - \cos \theta } \right)$$
$$ \Rightarrow $$ $${1 \over 2}\left( {{2 \over 3} + 1} \right){\left( {{{18} \over 5}} \right)^2} = \left( {m + M} \right)g \times {{M \times {l \over 2} + ml} \over {\left( {M + m} \right)}}\left( {1 - \cos \theta } \right)$$
$$ \Rightarrow $$ $${1 \over 2} \times {5 \over 3} \times {{18 \times 18} \over {25}} = 10 \times \left( {2 \times {1 \over 2} + 1} \right)\left( {1 - \cos \theta } \right)$$
$$ \Rightarrow $$ $${5 \over 6} \times {{18 \times 18} \over {25}} = 20\left( {1 - \cos \theta } \right)$$
$$ \Rightarrow $$ $$\left( {1 - \cos \theta } \right) = {{18} \over 5} \times {3 \over {20}}$$
$$ \Rightarrow $$ $${\cos \theta = 1 - {{27} \over {50}}}$$
$$ \Rightarrow $$ $${\cos \theta = {{23} \over {50}}}$$
$$ \Rightarrow $$ $$\theta $$ $$ \simeq $$ 63o
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