A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance. For the next loading, the spring is compressed a distance. How much faster does the second dart leave the gun compared with the first

Complete question is;

A dart is inserted into a spring - loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is compressed a distance 2x. How much faster does the second dart leave the gun compared to the first?

Answer:

The second dart leaves the gun two times faster than the first one.

Explanation:

If we assume there was no energy loss during the spring - dart energy transfer, we can easily apply the principle of conservation of energy. So;

Potential energy = kinetic energy

Thus;

½kx² = ½mv²

Making velocity "v" the subject, we have;

v = √ (kx²/m)

Since the initial distance is "x", thus initial launching velocity is;

v1 = √ (kx²/m)

Since next distance is 2x, thus, second launch velocity is;

v2 = √ (k (2x) ²/m)

Expanding, we have;

v2 = √ (4kx²/m)

v2 = 2√ (kx²/m)

Comparing this to the one gotten for v1 earlier, we can see that it is double v1.

So, v2 = 2v1

Hence, The second dart leaves the gun two times faster than the first one.