Question

When the spring on a toy gun is compressed by a distance \(x\), it will shoot a rubber ball straight up to a height of \(h .\) Ignoring air resistance, how high will the gun shoot the same rubber ball if the spring is compressed by an amount \(2 x ?\) Assume \(x<<h\)

Short answer

Answer: The ball will reach twice the original height, so h2 = 2h.

Step-by-step solution

Determine the work done on the ball and the initial potential energy of the spring

Calculate the initial potential energy of the spring when it is compressed by distance \(x\). According to the Hooke's Law, this potential energy (\(PE_{1}\)) can be represented as: \(PE_{1} = \frac{1}{2}kx^2\) Where \(k\) is the spring constant.

Calculate the gravitational potential energy at height h

When the rubber ball reaches the height \(h\), all the potential energy of the spring is transferred into gravitational potential energy. We can represent this gravitational potential energy (\(PE_{g}\)) as: \(PE_{g} = mgh\) Where \(m\) is the mass of the ball, and \(g\) is the acceleration due to gravity.

Relate the initial potential energy of the spring to the gravitational potential energy of the ball

Since the potential energy transferred from the spring to gravity, we can relate the initial potential energy of the spring to the gravitational potential energy of the ball: \(\frac{1}{2}kx^2 = mgh\)

Determine the height when the spring is compressed by 2x

Now, we need to find the height \(h_{2}\) when the spring is compressed by a distance \(2x\). Calculate the initial potential energy of the spring when it is compressed by distance \(2x\): \(PE_{2} = \frac{1}{2}k(2x)^2 = 2kx^2\)

Relate the potential energy of the spring compressed by 2x to the new gravitational potential energy

Set the potential energy of the compressed spring equal to the new gravitational potential energy \(PE_{g2}\): \(2kx^2 = mg(h_{2})\)

Solve for the new height h2

We now have two equations relating the potential energies: mgh = 1/2kx^2 and 2kx^2 = mgh2. Solve for \(h_{2}\): \(h_{2} = \frac{2kx^2}{mg}\) From the first equation, we can write the following: \(\frac{kx^2}{2mg} = h\) Multiply both sides of the equation by 2 to get: \(\frac{kx^2}{mg} = 2h\) Compare this result to the equation \(h_{2} = \frac{2kx^2}{mg}\) We can now see that \(h_{2} = 2h\), which means the ball will reach twice the original height when the spring is compressed by a distance \(2x\).