Combining Conservation Laws.? A 15.0-kg block is attached to a very light horizontal spring of force constant 500.0 N/m and is resting on a frictionless horizontal table (?Fig. E8.44) ? . Suddenly it is struck by a 3.00-kg stone traveling horizontally at 8.00 m/s to the right, whereupon the stone rebounds at 2.00 m/s horizontally to the left. Find the maximum distance that the block will compress the spring after the collision.

Solution 44E Step 1: Mass of stone m = 3.0 kg 1 Mass of block m = 152kg Spring constant k = 500.0 N/m Initial velocity of stone u = 810 m/s Initial velocity of block u = 0 m/s 2 After collision Stone rebounds with velocity v = -2.0m1 (opposite direction -x) Step 2: To find the maximum distance moved by the block Final velocity of block v is fou2d using conservation of momentum. Total momentum before collision is equal to the total momentum after collision. m u +m u = m v + m v ---(1) 1 1 2 2 1 1 2 2 Since initial velocity of the block is zero, (1) is changed as m 1 1m v1 1 v 2 2 Step 3: To find the velocity of block, the above equation is rearranged as follows m 1u 1v )1 v2 m 2 3*(8(2)) v2 15 v2 m/s The block will move with the velocity 2 m/s in the +x direction