Question

Can some one please help!!!!!!
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Answer 1

a) The launch velocity of the rocket is 5.48 m/s

b) The maximum height is 1.53 m

Explanation:

a)

We can solve this part by applying the law of conservation of energy, by considering the kinetic energy and the elastic potential energy only, since there is no change in gravitational potential energy and no friction is involved.

The total energy when the spring is compressed is:

$E=KE_i + PE_{si}$

with

$KE_i = 0$ (initial kinetic energy is zero)

$PE_{si} = \frac{1}{2}kx^2$ is the elastic potential energy stored in the spring, with

k = 450 N/m (spring constant)

x = 0.10 m (compression of the spring)

The total energy when the spring is relased is:

$E=KE_f + PE_{sf}$

with

$KE_f = \frac{1}{2}mv^2$ (final kinetic energy), with

m = 0.15 kg (mass of the rocket)

v = velocity of launch of the rocket

$PE_{sf} = 0$ (elastic potential energy is zero when the spring is released)

Combining the two equations we get

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

And solving for v,

$v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(450)(0.10)^2}{0.15}}=5.48 m/s$

b)

In this part instead we consider only the kinetic energy and the gravitational potential energy, since the spring is at rest so its energy is now zero.

The total energy at the launch is:

$E=KE_i + PE_{gi}$

where

$KE_i = \frac{1}{2}mv^2$ (initial kinetic energy), with

m = 0.15 kg (mass of the rocket)

v = 5.48 m/s (velocity of launch of the rocket)

$PE_{gi}=0$ (initial gravitational potential energy is zero)

The total energy at the point of maximum height is:

$E=KE_f + PE_{gf}$

where

$KE_f = 0$ (kinetic energy is zero since speed is zero)

$PE_{gf}=mgh$ (final gravitational potential energy), with

m = 0.15 kg

$g=9.8 m/s^2$ (acceleration of gravity)

h = ? (maximum height)

Combining the two equations we find

$\frac{1}{2}mv^2 = mgh$

And solving for h,

$h=\frac{v^2}{2g}=\frac{(5.48)^2}{2(9.8)}=1.53 m$

Learn more about potential energy and kinetic energy:

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