Question

A spring gun shoots out a plastic ball at speed v0. The spring is then compressed twice the distance it was on the first shot. By what factor is the ball’s speed increased? Explain.

Short answer

The distance travelled by an object per unit time is known as speed.

Step-by-step solution

Introduction 

A divisor of an integer n, also known as a factor of n, is an integer m that may be multiplied by another integer to give the result n. One may also state that n is a multiple of m in this scenario.

Explanation 

Explain by what factor the speed of the ball is increased.

Use the law of conservation of energy for the spring when it shoots out first the plastic ball at a speed

$\frac{1}{2}k{x}^2=\frac{1}{2}m{v}_0^2$

Here, $k$is the spring constant, $x$is the initial compression of the spring, $m$is the mass of the plastic ball and $v_o$is the initial speed of the plastic ball.

Rearrange the above equation for $v_o$.

$k{x}^2=m{v}_0^2$

$v_o=\sqrt{\frac{k{x}^2}{m}}\dots ..\left(1\right)$

Use the law of conservation of energy for the spring and ball system when the spring shoot out the plastic ball again and in this case the compression is twice the distance it was on the first shot.

$\frac{1}{2}k{x}^2=\frac{1}{2}m{v}_0^{\text{'}2}$

Here, $x^{\text{'}}$is the final compression and $v_0^{\text{'}}$is the final speed of the plastic ball.

Rearrange the above equation for $v_0^{\text{'}}$.

$k{x}^{\text{'}2}=m{v}_0^{\text{'}2}$

$v_0^{\text{'}}=\sqrt{\frac{k{x}^{\text{'}2}}{m}}\dots \dots \left(2\right)$

In the second shot of the ball, the spring compressed twice the distance it was on the first shot, that is,

$x^{\text{'}}=2x$

Substitute $2x$for $x^{\text{'}}$in above equation $\left(2\right)$.

$v_0^{\text{'}}=\sqrt{\frac{k(2x{)}^2}{m}}$

$=\sqrt{\frac{4k{x}^2}{m}}$

$=\left(2\right)\left(\sqrt{\frac{k{x}^2}{m}}\right)$

Substitute the equation$\left(1\right)$ in the above equation.

$v_0^{\text{'}}=2v_0$

Therefore, the speed of the plastic ball increases by $2$times the initial speed.