Question

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)

Answer 1

a)
`v=13.2171\,m.s^(-1)`

b)
`H=8.9605\,m`

Step-by-step explanation:

Given:

mass of bullet,
`m=4.97* 10^(-3)\,kg`

compression of the spring,
`\Delta x=0.0476\,m`

force required for the given compression,
`F=9.12 \,N`

(a)

We know

`F=m.a`

where:

a= acceleration

`9.12=4.97* 10^(-3)* a`

`a\approx 1835\,m.s^(-2)\\`

we have:

initial velocity,
`u=0\,m.s^(-1)`

Using the eq. of motion:

`v^2=u^2+2a.\Delta x`

where:

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

`v^2= 0^2+2* 1835* 0.0476`

`v=13.2171\,m.s^(-1)`

(b)

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

so,

`u=13.2171\,m.s^(-1)`

∵At maximum height the final velocity will be zero

`v=0\,m.s^(-1)`

Using the equation of motion:

`v^2=u^2-2g.h`

where:

h= height

g= acceleration due to gravity

`0^2=13.2171^2-2* 9.8* h`

`h=8.9129\,m`

is the height from the release position of the spring.

So, the height from the latched position be:

`H=h+\Delta x`

`H=8.9129+0.0476`

`H=8.9605\,m`