7. a bullet of mass 10g strikes a ballistic pendulum of mass 2kg. the center of mass of the pendulum rises a vertical distance of 12cm. assuming that the bullet remains embedded in the pendulum, calculate the bullet’s initial speed.

m = mass of the bullet = 10 g = 10 x 10⁻³ kg = 0.01 kg       (since 1 g = 10⁻³ kg)

M = mass of pendulum = 2 kg

h = height to which pendulum rises = 12 cm = 0.12 m

V = velocity of the pendulum-bullet combination after collision = ?

Using conservation of energy

kinetic energy of the combination just after collision = Potential energy gained due to raise in height of the center of mass

(0.5) (m + M) V² = (m + M) gh

V = sqrt(2gh)

inserting the values

V = sqrt(2 x 9.8 x 0.12)

V = 1.5 m/s

v = velocity of the bullet before the collision

using conservation of momentum

momentum of bullet before collision = momentum of bullet-pendulum combination after collision

m v = (m + M) V

(0.01) v = (0.01 + 2) (1.5)

v = 301.5 m/s

hence initial speed of the bullet is 301.5 m/s

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.