Question

base your answer to this question on the information below and on your knowledge of physics. A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. Newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere. The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched.

Answer 1

Answer: v = 2.24 m/s

Explanation: The Law of Conservation of Energy states that total energy is constant in any process and, it cannot be created nor destroyed, only transformed.

So, in the toy launcher, the energy of the compressed spring, called Elastic Potential Energy (PE), transforms into the movement of the plastic sphere, called Kinetic Energy (KE). Since total energy must be constant:

$KE_{i}+PE_{i}=KE_{f}+PE_{f}$

where the terms with subscript i are related to the initial of the process and the terms with subscript f relates to the final process.

The equation is calculated as:

$\frac{1}{2}kx^{2}+0=0+\frac{1}{2}mv^{2}$

$\frac{1}{2}kx^{2}=\frac{1}{2}mv^{2}$

$\frac{1}{2}50(0.1)^{2}=\frac{1}{2}(0.1)v^{2}$

$v^{2}=\frac{50(0.1)^{2}}{0.1}$

$v=\sqrt{50(0.1)}$

$v=\sqrt{5}$

v = 2.24

The maximum speed the plastic sphere will be launched is 2.24 m/s.

Answer 2

The maximum speed of plastic sphere is 2.236 m/s, when sphere is launched.

The maximum velocity of sphere can be calculated by,

$V_{max}= x \sqrt {\dfrac km}$

Where,

$V_{max}$ - maximum velocity of the sphere

$x$ - length of stretch = 0.10 m

$k$-  spring constant = 50 N

$m$- mass of sphere =  0.10 kg

Put the values in the equation,

$V_{max}= 0.1\sqrt {\dfrac {50}{0.1}}\\\\V_{max}= 0.1\times {22.36}\\\\V_{max}= 2.236$

Therefore,  the maximum speed of plastic sphere is 2.236 m/s, when sphere is launched.

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