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

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down