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

To solve this problem, we can use the conservation of momentum and the given information about the kinetic energy loss during the collision.<br /><br />Given information:<br />- Mass of the block: 4 kg<br />- Initial speed of the block: 10 m/s<br />- Mass of the stationary box: 2 kg<br />- Kinetic energy loss during the collision: 25%<br /><br />Step 1: Calculate the initial kinetic energy of the block.<br />Initial kinetic energy = (1/2) × mass of block × (initial speed of block)^2<br />Initial kinetic energy = (1/2) × 4 kg × (10 m/s)^2 = 200 J<br /><br />Step 2: Calculate the kinetic energy after the collision.<br />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.<br />Kinetic energy after collision = 0.75 × Initial kinetic energy = 0.75 × 200 J = 150 J<br /><br />Step 3: Use the conservation of momentum to find the final speed of the block and the box after the collision.<br />Let the final speed of the block be v1 and the final speed of the box be v2.<br />Momentum before collision = Momentum after collision<br />Mass of block × Initial speed of block = (Mass of block × Final speed of block) + (Mass of box × Final speed of box)<br />4 kg × 10 m/s = (4 kg × v1) + (2 kg × v2)<br /><br />Step 4: Use the kinetic energy formula to find the final speed of the block and the box.<br />Kinetic energy after collision = (1/2) × Mass of block × (Final speed of block)^2 + (1/2) × Mass of box × (Final speed of box)^2<br />150 J = (1/2) × 4 kg × (v1)^2 + (1/2) × 2 kg × (v2)^2<br /><br />Step 5: Solve the equations from Step 3 and Step 4 to find the final speed of the block and the box.<br />Solving the equations, we get:<br />v1 = 3.88 m/s<br />v2 = 5.61 m/s<br /><br />Therefore, the correct answer is:<br />a. $3.88m/s$