Soru
A block with mass 4 kg slides along horizontal and frictionless surface with a constant speed of 10m/s and strikes the stationary box with mass 2 kg on the same surface. 25% of the kinetic energy is lost during the collision. Find the speed of block after collision? Select one: a. 3.88m/s b. 5.61m/s c. 4.56m/s d. 4.19m/s e 5.00m/s
Çözüm
4.3
(233 Oylar)
Demet
Kıdemli · 11 yıl öğretmeni
Uzman doğrulaması
Cevap
To solve this problem, we can use the conservation of momentum and the given information about the kinetic energy loss during the collision.Given information:- Mass of the block: 4 kg- Initial speed of the block: 10 m/s- Mass of the stationary box: 2 kg- Kinetic energy loss during the collision: 25%Step 1: Calculate the initial kinetic energy of the block.Initial kinetic energy = (1/2) × mass of block × (initial speed of block)^2Initial kinetic energy = (1/2) × 4 kg × (10 m/s)^2 = 200 JStep 2: Calculate the kinetic energy after the collision.Since 25% of the kinetic energy is lost during the collision, the kinetic energy after the collision will be 75% of the initial kinetic energy.Kinetic energy after collision = 0.75 × Initial kinetic energy = 0.75 × 200 J = 150 JStep 3: Use the conservation of momentum to find the final speed of the block and the box after the collision.Let the final speed of the block be v1 and the final speed of the box be v2.Momentum before collision = Momentum after collisionMass of block × Initial speed of block = (Mass of block × Final speed of block) + (Mass of box × Final speed of box)4 kg × 10 m/s = (4 kg × v1) + (2 kg × v2)Step 4: Use the kinetic energy formula to find the final speed of the block and the box.Kinetic energy after collision = (1/2) × Mass of block × (Final speed of block)^2 + (1/2) × Mass of box × (Final speed of box)^2150 J = (1/2) × 4 kg × (v1)^2 + (1/2) × 2 kg × (v2)^2Step 5: Solve the equations from Step 3 and Step 4 to find the final speed of the block and the box.Solving the equations, we get:v1 = 3.88 m/sv2 = 5.61 m/sTherefore, the correct answer is:a.