A spring of force constant k = 3000 N/m launches 1-kg marbles upward into a 20-kg block, as shown at right. The spring is compressed 0.2 m from its natural length and then released. The marble travels upward 0.2 m as the spring pushes it and then travels upward an additional 0.8 m to strike the bottom of the block. The block then flies upward into the air.

Question

anonymous · Answer

What is the speed of the marble as it hits the block?

anonymous · Answer

energy conservation.
state one is when the spring is completely compressed.
consider the system of the spring and the marble.
State 1 Energy: 1/2 k x^2 + mgh, define this point as h=0, so it's just 1/2 *3000*0.2^2 = 40 J
Now when the spring is unstretched we have
40 = mgh + 1/2 mv^2 = 1(9.81)(0.2) + 1/2 (1) (v)^2
v of ball when leaving = 8.722 m/s

Keep going...
mgh + 1/2 mv^2 = mgh + 1/2 mv^2
(1)(9.81) (0.1) + 1/2 (1) (8.722)^2 = (1)(9.81)(1) + 1/2 (1) v^2 
v right before hitting = 7.64 m/s
hoepfully no calculator mistakes there.. hope that helps