An 8.00 gram bullet is fired horizontally into a 0.992 kg block of wood sitting

Question

An 8.00 gram bullet is fired horizontally into a 0.992 kg block of wood sitting on the edge of a 1.50 m tall table. The bullet becomes embedded in the block and together they travel as a projectile off the table and to a horizontal distance of 1.80 m from the base of the table.
A) Calculate the velocity of the block and bullet system as it leaves the edge of the table.
B) Determine the speed of the bullet just before striking the block.
C) If the bullet went through the wood and escaped with a speed of 1/4 of the value found in part B, how far horizontally would the wood block travel from the base of the table?

Explanation / Answer

Mass of the bullet m = 8 g = 0.008 Kg

Mass of the block M = 0.992 Kg

Height of the table h = 1.5 m

Horizontal distance travelled R = 1.80 m

(A)

Time of flight T = Sqrt(2h/g) = 0.553 s

Velocity of the bullet and block system v = R/T = 1.80/0.553 = 3.254 m/s

(B)

Velocity of the bullet just before striking the block u = [(m + M)/m]v

                                                                             = [1/0.008]3.254

                                                                             = 406.75 m/s

(C)

According to the law of conservation of momentum

                              mu + 0 = m(u/4) + Mv'

                                    Mv' = 3mu/4

                                       v' = 3mu/4M

                                          = 2.46 m/s

Horizontal distance covered by the block R' = v' T = 2.46 * 0.553 = 1.36 m

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