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.
