Answer:
(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
