A Hooke's law spring is compressed a distance \( d \) and is used to launch a mass \( m \) vertically to a height \( h \) above its starting position. Under the same compression \( d \) the spring is now used to launch a mass of \( 2 m \). How high does this second mass rise? 1) \( h \) 2) \( h / 2 \) 3) \( h / 1.41 \) 4) \( h / 4 \) 5) \( h / 8 \)

When the spring is compressed, it stores elastic potential energy, which gets converted into gravitational potential energy when launching the mass. The energy stored in the spring can be given by \( \frac{1}{2} k d^2 \), where \( k \) is the spring constant. For the first mass \( m \), the energy converts to gravitational potential energy \( mgh \). When the second mass \( 2m \) is launched with the same compression, the energy available is still \( \frac{1}{2} k d^2 \), which now must equal \( 2mgH \) for some new height \( H \). Setting these equal gives \( mgH = \frac{1}{2} k d^2 \). As you can see, the height \( H \) is halved for the second mass, resulting in \( H = h / 2 \). Therefore, the height to which the second mass \( 2m \) rises is \( h / 2 \), which corresponds to answer choice 2.