6. A .25 kg arrow with a velocity of 12 m/s to the west strikes and pierces the center of a 6.8 kg target. a. What is the final velocity of the combined mass? b. What is the decrease in kinetic energy during the collision?

(a) the final velocity of the combined mass is 9.43 m/s

(b) the decrease in kinetic energy during the collision is 386.1 J

Explanation:

Given;

mass of arrow, m₁ = 25 kg

initial velocity of arrow, u₁ = 12 m/s

mass of target, m₂ = 6.8 kg

initial velocity of the target, u₂ = 0

Part (a)

From the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁+m₂)

where;

v is the final velocity of the combined mass

25 x 12 + 0 = v(25 + 6.8)

300 = v(31.8)

v = 300/31.8

v = 9.43 m/s

Part(b)

Kinetic Energy, K.E = ¹/₂mv²

Initial kinetic energy =  ¹/₂m₁u₁² + ¹/₂m₂u₂²  = ¹/₂ x 25 x (12)² + 0 = 1800 J

Final kinetic energy = ¹/₂m₁v² + ¹/₂m₂v² = ¹/₂v²(m₁ + m₂)

                                                               = ¹/₂ x (9.43)²(25+6.8)

                                                               = 1413.91 J

Decrease in kinetic energy = Initial K.E - Final K.E

Decrease in kinetic energy = 1800J - 1413.91 J = 386.1 J