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 center 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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a
821.4m/s
b
811.4m/s
c
841.4m/s
d
831.4m/s
answer is D.
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Detailed Solution
Conservation of momentumPi=Pf ⇒0.01 ×u+0 =6×v ⇒v=0.01u6Using energy conservation , after collision ⇒12×6×u6002=6×9.8×9.8×10−2⇒u=6×98×2=5882 m/s=831.4 m/s
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 center 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)