Question

A toy gun uses a spring to shoot plastic balls (m = 50 g). The spring is compressed by 3.0 cm. Let k=2.22 × 105 N/m. (a) Of course, you have to do some work on the gun to arm it. How much work do you have to do? (b) Suppose you fire the gun horizontally. How fast does the ball leave the gun? (c) Now suppose you fire the gun straight upwards. How high does the ball go?

Answer 1

Answer:

Explanation:

Given that:

mass of the plastic ball = 50 g = 50 × 10⁻³ kg

spring constant (k) = 2.22 × 10⁵ N/m

compression of spring (x) = 3.0 cm = 3 × 10⁻² m

a) Work done:

$= \dfrac{1}{2}kx^2 \\ \\ = \dfrac{1}{2}\times 2.22 \times 10^{5}\times (3 \times 10^{-2})^2 \\ \\ = 99.9 \ J$

b) When the gun is fired horizontally;

In the spring, the potential energy is changed to kinetic energy.

i.e.

$\implies \dfrac{1}{2}kx^2 = \dfrac{1}{2}mv^2 \\ \\ 2.22 \times 10^5 \times (3\times 10^{-2}) ^2 =50 \times 10^{-3} \times v^2\\ \\ \mathbf{v= 63.2 m/s}$

c) The max height the bullet will reach if the gun is being shot upward is:

$H_{max}= \dfrac{v^2}{2g} \\ \\ H_{max}= \dfrac{63.2^2}{2\times 9.8}$

$\mathbf{H_{max}= 203.9 \ m}$