Soru
EXAMPLE: A ball of mass 0.440 kg moving east +x direction)with a speed of 3,80 m/s collides head-on with a 0.220-kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?
Çözüm
3.3
(178 Oylar)
Gokhan
Usta · 5 yıl öğretmeni
Uzman doğrulaması
Cevap
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.Given information:- Mass of the first ball (m1) = 0.440 kg- Velocity of the first ball (v1) = 3.80 m/s (eastward)- Mass of the second ball (m2) = 0.220 kg- Velocity of the second ball (v2) = 0 m/s (at rest)Step 1: Calculate the total momentum before the collision.Total momentum before the collision = m1 × v1 + m2 × v2Total momentum before the collision = 0.440 kg × 3.80 m/s + 0.220 kg × 0 m/sTotal momentum before the collision = 1.672 kg·m/s (eastward)Step 2: Calculate the total kinetic energy before the collision.Total kinetic energy before the collision = (1/2) × m1 × v1^2 + (1/2) × m2 × v2^2Total kinetic energy before the collision = (1/2) × 0.440 kg × (3.80 m/s)^2 + (1/2) × 0.220 kg × (0 m/s)^2Total kinetic energy before the collision = 2.624 JStep 3: Calculate the final velocities of the balls after the collision.Since the collision is perfectly elastic, the total momentum and total kinetic energy will be conserved.Let's denote the final velocity of the first ball as v1' and the final velocity of the second ball as v2'.Using the conservation of momentum:m1 × v1 + m2 × v2 = m1 × v1' + m2 × v2'1.672 kg·m/s = 0.440 kg × v1' + 0.220 kg × v2' (1)Using the conservation of kinetic energy:(1/2) × m1 × v1^2 + (1/2) × m2 × v2^2 = (1/2) × m1 × v1'^2 + (1/2) × m2 × v2'^22.624 J = (1/2) × 0.440 kg × v1'^2 + (1/2) × 0.220 kg × v2'^2 (2)Solving equations (1) and (2) simultaneously, we get:v1' = 2.80 m/s (eastward)v2' = -1.40 m/s (westward)Therefore, after the collision, the first ball will move eastward with a speed of 2.80 m/s, and the second ball will move westward with a speed of 1.40 m/s.