Soru
Question 1 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
4.1
(216 Oylar)
İsmail
Usta · 5 yıl öğretmeni
Uzman doğrulaması
Cevap
To solve this problem, we can use the conservation of momentum and the given information about the loss of kinetic energy.Given:- Mass of the block (m1) = 4 kg- Initial speed of the block (v1) = 10 m/s- Mass of the stationary box (m2) = 2 kg- Percentage of kinetic energy lost during the collision = 25%Step 1: Calculate the initial kinetic energy of the block.Initial kinetic energy (KE1) = (1/2) * m1 * v1^2KE1 = (1/2) * 4 kg * (10 m/s)^2KE1 = 200 JStep 2: Calculate the kinetic energy lost during the collision.Kinetic energy lost = 25% of KE1Kinetic energy lost = 0.25 * 200 JKinetic energy lost = 50 JStep 3: Calculate the final kinetic energy of the block and the box after the collision.Final kinetic energy (KE2) = Initial kinetic energy - Kinetic energy lostKE2 = 200 J - 50 JKE2 = 150 JStep 4: Use the conservation of momentum to find the final speed of the block and the box after the collision.Total momentum before collision = Total momentum after collisionm1 * v1 = (m1 * v1') + (m2 * v2')4 kg * 10 m/s = (4 kg * v1') + (2 kg * v2')40 kg*m/s = 4 kg * v1' + 2 kg * v2'Step 5: Use the conservation of energy to find the final speed of the block and the box after the collision.KE2 = (1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2150 J = (1/2) * 4 kg * v1'^2 + (1/2) * 2 kg * v2'^2Now, we can solve these two equations simultaneously to find the final speed of the block (v1').After solving the equations, we get:v1' ≈ 3.88 m/sTherefore, the correct answer is:a.