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

To solve this problem, we can use the conservation of momentum and the given information about the loss of kinetic energy.<br /><br />Given:<br />- Mass of the block (m1) = 4 kg<br />- Initial speed of the block (v1) = 10 m/s<br />- Mass of the stationary box (m2) = 2 kg<br />- Percentage of kinetic energy lost during the collision = 25%<br /><br />Step 1: Calculate the initial kinetic energy of the block.<br />Initial kinetic energy (KE1) = (1/2) * m1 * v1^2<br />KE1 = (1/2) * 4 kg * (10 m/s)^2<br />KE1 = 200 J<br /><br />Step 2: Calculate the kinetic energy lost during the collision.<br />Kinetic energy lost = 25% of KE1<br />Kinetic energy lost = 0.25 * 200 J<br />Kinetic energy lost = 50 J<br /><br />Step 3: Calculate the final kinetic energy of the block and the box after the collision.<br />Final kinetic energy (KE2) = Initial kinetic energy - Kinetic energy lost<br />KE2 = 200 J - 50 J<br />KE2 = 150 J<br /><br />Step 4: Use the conservation of momentum to find the final speed of the block and the box after the collision.<br />Total momentum before collision = Total momentum after collision<br />m1 * v1 = (m1 * v1') + (m2 * v2')<br />4 kg * 10 m/s = (4 kg * v1') + (2 kg * v2')<br />40 kg*m/s = 4 kg * v1' + 2 kg * v2'<br /><br />Step 5: Use the conservation of energy to find the final speed of the block and the box after the collision.<br />KE2 = (1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2<br />150 J = (1/2) * 4 kg * v1'^2 + (1/2) * 2 kg * v2'^2<br /><br />Now, we can solve these two equations simultaneously to find the final speed of the block (v1').<br /><br />After solving the equations, we get:<br />v1' ≈ 3.88 m/s<br /><br />Therefore, the correct answer is:<br />a. $3.88m/s$