Question

spring-loaded toy gun projects a 5.9 g nerf pellet horizontally. The spring constant is 8.5 N/m, the barrel of the gun is 17 cm long, and a constant frictional force of 0.037 N exists between the barrel and the nerf pellet. If the spring is compressed 6.8 cm for this launch, determine the speed (in m/s) of the pellet as it leaves the barrel. (Assume the pellet is in contact with the barrel for the full length of the barrel.) m/s

Answer 1

Answer:

Solution:

As per the question:

Mass of the pellet, m = 5.9 g = 0.059 kg

Spring constant, k = 8.5 N/m

Length of the barrel, l = 17 cm = 0.17 m

Frictional force, F = 0.037 N

Compression in spring, $\Delta x = 6.8\ cm = 0.068\ m$

Now,

To calculate the speed of the pellet:

Using the principle of conservation of energy:

Change in spring potential energy is used in doing work against the friction force and provides the required kinetic energy:

$\frac{1}{2}k\Delta x^{2} = \frac{1}{2}mv^{2} + Fl$

$\frac{1}{2}\times 8.5\times 0.0068^{2} = \frac{1}{2}\times 5.9\times 10^{- 3}\times v^{2} + 0.037\times 0.17$

$v^{2} = 0.2797$

v = 0.5289 m/s

Answer 2

Answer:

$v=5.5836\ m.s^{-1}$

Explanation:

Given:

• mass of pellet, $m=5.9\times 10^{-3}\ kg$

• spring constant of the gun, $k=8.5\ N.m^{-1}$

• length of the barrel, $l=0.17\ m$

• frictional force offered by the barrel, $f=0.037\ N$

• compression of the spring, $\delta x=0.068\ m$

Spring force on the pellet:(inside the barrel)

$F=k.\delta x$

$F=8.5\times 0.068$

$F=0.578\ N$

Net force on the pellet:(INSIDE THE BARREL)

$F_n=F-f$

$F_n=0.578-0.037$

$F_n=0.541\ N$

Acceleration of the pellet:(inside the barrel)

$a=\frac{F_n}{m}$

$a=\frac{0.541}{0.0059}$

$a=91.6949\ m.s^{-2}$

Using the equation of motion:

$v^2=u^2+2a.l$

where, v & u are the final and initial velocities respectively

$v^2=0^2+2\times 91.6949\times 0.17$

$v=5.5836\ m.s^{-1}$ is the velocity at the exit of the barrel