JEE MAIN - Physics (2021 - 16th March Evening Shift - No. 5)
A large block of wood of mass M = 5.99 kg is hanging from two long massless cords. A bullet of mass m = 10 g is fired into the block and gets embedded in it. The (block + bullet) then swing upwards, their centre of mass rising a vertical distance h = 9.8 cm before the (block + bullet) pendulum comes momentarily to rest at the end of its arc. The speed of the bullet just before collision is :
(take g = 9.8 ms-2)
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(take g = 9.8 ms-2)
831.4 m/s
811.4 m/s
841.4 m/s
821.4 m/s
Explanation
$${P_i} = 0.01 \times u + 0 = {P_f} = 6 \times v$$
$$v = {{0.01u} \over 6}$$
using energy conservation
$${1 \over 2} \times 6 \times {\left( {{u \over {600}}} \right)^2} = 6 \times 9.8 \times 9.8 \times {10^{ - 2}}$$
$$ \Rightarrow $$ $$u = 6 \times 98 \times \sqrt 2 = 588\sqrt 2 $$ m/s
$$v = {{0.01u} \over 6}$$
using energy conservation
$${1 \over 2} \times 6 \times {\left( {{u \over {600}}} \right)^2} = 6 \times 9.8 \times 9.8 \times {10^{ - 2}}$$
$$ \Rightarrow $$ $$u = 6 \times 98 \times \sqrt 2 = 588\sqrt 2 $$ m/s
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