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A spring gun is made by compressing a spring in a tube and then latching the spring at

the compressed position. A 4.97-g pellet is placed against the compressed and latched

spring. The spring latches at a compression of 4.76 cm, and it takes a force of

9.12 N to compress the spring to that point.

(a) If the gun is fired vertically, how fast (m/s) is the pellet moving when it loses contact

with the spring?

(b) To what maximum height (m) will the pellet rise? (as measured from the original

latched position)

Respuesta :

Answer:

a)[tex]v=13.2171\,m.s^{-1}[/tex]

b)[tex]H=8.9605\,m[/tex]

Explanation:

Given:

mass of bullet, [tex]m=4.97\times 10^{-3}\,kg[/tex]

compression of the spring, [tex]\Delta x=0.0476\,m[/tex]

force required for the given compression, [tex]F=9.12 \,N[/tex]

(a)

We know

[tex]F=m.a[/tex]

where:

a= acceleration

[tex]9.12=4.97\times 10^{-3}\times a[/tex]

[tex]a\approx 1835\,m.s^{-2}\\[/tex]

we have:

initial velocity,[tex]u=0\,m.s^{-1}[/tex]

Using the eq. of motion:

[tex]v^2=u^2+2a.\Delta x[/tex]

where:

v= final velocity after the separation of spring with the bullet.

[tex]v^2= 0^2+2\times 1835\times 0.0476[/tex]

[tex]v=13.2171\,m.s^{-1}[/tex]

(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

[tex]u=13.2171\,m.s^{-1}[/tex]

∵At maximum height the final velocity will be zero

[tex]v=0\,m.s^{-1}[/tex]

Using the equation of motion:

[tex]v^2=u^2-2g.h[/tex]

where:

h= height

g= acceleration due to gravity

[tex]0^2=13.2171^2-2\times 9.8\times h[/tex]

[tex]h=8.9129\,m[/tex]

is the height from the release position of the spring.

So, the height from the latched position be:

[tex]H=h+\Delta x[/tex]

[tex]H=8.9129+0.0476[/tex]

[tex]H=8.9605\,m[/tex]

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