The 6.0 kg block is at rest while the 2.0 kg block moves at 1.0 m/s towards the 6.0 kg block, from an initial distance of 16 cm from it. The coefficient of kinetic friction between the blocks and the ground is 0.20. The collision between the blocks is perfectly elastic. The separation between the blocks, (in mm), after both the blocks come to rest is 9K. Find the value of K. Take g=10 m/s².
Physics · Mixed Chapter
JEE Mains

Question

The 6.0 kg block is at rest while the 2.0 kg block moves at 1.0 m/s towards the 6.0 kg block, from an initial distance of 16 cm from it. The coefficient of kinetic friction between the blocks and the ground is 0.20. The collision between the blocks is perfectly elastic. The separation between the blocks, (in mm), after both the blocks come to rest is 9K. Find the value of K. Take g=10 m/s².

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Kinetic friction gives retardation μg=2 m/s². The 2 kg block travels 0.16 m before collision, so impact speed satisfies v²=1²−2(2)(0.16)=0.36, i.e. v=0.6 m/s. Step 2: For a perfectly elastic head-on collision with stationary 6 kg block: v₁′=[(2−6)/(2+6)]0.6=−0.3 m/s and v₂′=[2×2/(2+6)]0.6=+0.3 m/s. Step 3: Both blocks decelerate at 2 m/s² and each travels s=(0.3)²/(2×2)=0.0225 m in opposite directions. Step 4: Their final separation is 2s=0.045 m=45 mm=9K mm. Hence K=5.