A slingshot obeying Hooke’s law is used to launch pebbles vertically into the ai

Question

A slingshot obeying Hooke’s law is used to launch pebbles vertically into the air. You observe that if you pull a pebble back 23.5 cm against the elastic band, the pebble goes 6.6 m high.

a) Assuming that air drag is negligible, how high will the pebble go if you pull it back 47.0 cm instead?

Express your answer to two significant figures and include the appropriate units.

b) How far must you pull it back so it will reach 13.2 m ?

Express your answer to three significant figures and include the appropriate units.

c) If you pull a pebble that is twice as heavy back 23.5 cm , how high will it go?

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

given,

if we pull back a pebble back 23.5 cm against the elastic band the pebble goes 6.6m high

sicne the energy remains conserved in this process so,

0.5 * k * x^2 = mgh

0.5 * k * 0.235^2 = m * 9.8 * 6.6

k = m * 2342.42

Now if elastic band is pulled back to 47 cm then

0.5 * k * 0.47^2 = m * 9.8 * h

since k = m * 2342.42

0.5 * m * 2342.42 * 0.47^2 = m * 9.8 * h

h = 26.4 m

a) pebble will go 26.40 m high

if pebble have to reach 13.2 m then

0.5 * m * 2342.42 * x^2 = m * 9.8 * 13.2

x = 0.33234 or 33.234 cm

b) elastic band have to pull 33.234 cm back

Let m' = 2*m, or m = 0.5*m' so,

0.5 * k * 0.235^2 = 2m * g * h

Since k = m * 2342.42 Thus,

0.5 * m * 2342.42 * 0.235^2 = 2m * g * h

h = 3.3 m

c) if the pebble is twice as heavy then it will go till 3.3 m

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