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 = 9.8ms?²)
Using conservation of linear momentum, we can write P? = P? ⇒ mv = (M+m)V Using conservation of Mechanical energy, we can write ½ (M+m)V² = (M+m)gh ⇒ V = √2gh ⇒ v = (M+m)/m)√2gh = (6/0.01)√ (2×9.8×0.098) = 600 × 1.386 = 831.4 m/s
<p>Using conservation of linear momentum, we can write<br>P? = P? ⇒ mv = (M+m)V<br>Using conservation of Mechanical energy, we can write<br>½ (M+m)V² = (M+m)gh ⇒ V = √2gh<br>⇒ v = (M+m)/m)√2gh<br>= (6/0.01)√ (2×9.8×0.098) = 600 × 1.386 = 831.4 m/s</p>
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