JEE MAIN - Physics (2019 - 10th January Morning Slot - No. 11)
A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is - (g = 10 ms–2)
30 m
40 m
20 m
10 m
Explanation
_10th_January_Morning_Slot_en_11_1.png)
Time taken for the particles to collide,
t = $${f \over {{V_{rel}}}} = {{100} \over {100}} = 1$$ sec
Speed of wood just before collision = gt = 10 m/s
& speed of bullet just before collision v-gt
= 100 $$-$$ 10 = 90 m/s
Now, conservation of linear momentum just before and after the collision -
$$-$$ (0.02) (1v) + (0.02) (9v) = (0.05)v
$$ \Rightarrow $$ 150 = 5v
$$ \Rightarrow $$ v = 30 m/s
Max. height reached by body h = $${{{v^2}} \over {2g}}$$
_10th_January_Morning_Slot_en_11_2.png)
h = $${{30 \times 30} \over {2 \times 10}}$$ = 45m
$$ \therefore $$ Height above tower = 40 m
Comments (0)
