Question

a spring gun shoots out a plastic ball at speed v. the spring is then compressed twice the distance it was on the first shot. by what factor is the ball's speed increased?

Answer 1

Answer:√4 = 2.

Step-by-step explanation:

When the spring gun is fired for the first time, let's assume that the spring was compressed by a distance of x, and the plastic ball is shot out at a speed of v.

Now, the spring is compressed twice the distance it was on the first shot. Therefore, the new compression distance is 2x.

Let's assume that the speed of the plastic ball after the second shot is v'. We can use the principle of conservation of energy to relate the speed of the ball to the compression distance of the spring.

For the first shot, the energy stored in the spring (½kx², where k is the spring constant) is converted into the kinetic energy of the ball (½mv², where m is the mass of the ball). Therefore,

½kx² = ½mv²

For the second shot, the energy stored in the spring (½k(2x)² = 2kx²) is again converted into the kinetic energy of the ball (½mv'²). Therefore,

2kx² = ½mv'²

Dividing the second equation by the first equation, we get:

v'²/v² = 4

Therefore, the speed of the plastic ball is increased by a factor of √4 = 2. So, the ball's speed is increased by a factor of 2.

Answer 2

The speed of the plastic ball is increased by a factor of the square root of 2 (approximately 1.41).

To find the factor by which the speed of the plastic ball is increased when the spring is compressed twice the distance, we can use the principle of conservation of energy.

When the spring is compressed twice the distance, the potential energy stored in the spring is doubled. This means that the kinetic energy of the plastic ball will also be doubled, as energy is conserved.

Since the kinetic energy is directly related to the square of the speed, the speed of the ball will be increased by a factor of the square root of 2 (approximately 1.41).