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3 years ago

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William Tell shoots an apple from his son's head. The speed of the 100-g arrow just before it strikes the apple is 29.4 m/s, and

at the time of impact it is traveling horizontally. If the arrow sticks in the apple and the arrow/apple combination strikes the ground 8.50 m behind the son's feet, how massive was the apple

1 answer:

Cerrena [4.2K]3 years ago

6 0

To solve this problem it is necessary to apply the kinematic equations of motion (vertical in this case) as well as the momentum conservation equations.

For conservation of the moment we have to

$m_1v_1 = (m_1+m_2)v_f$

Where

$m_{1,2} =$ Mass of each object

$v_1 =$ Initial velocity of arrow

$v_f$ = Final Velocity

Time in flight for arrow-apple combination is

$t= \sqrt{\frac{2h}{g}}$

Where,

h = Max height

g = Gravitational acceleration

Now after the impact arrow-apple combination have horizontal velocity V and it is

$V= \frac{x}{t}$

From the previous definition we have that the value of time would be,

$V = \frac{x}{\sqrt{\frac{2h}{g}}}$

Assuming the son's height is 1.85m, then we should

$V = \frac{8.5}{\sqrt{\frac{2(1.85)}{9.8}}}$

$V = 13.83m/s$

Applying again the conservation equation we can obtain the value of the apple mass as:

$m_2 = m_1(\frac{V_1}{ V_f} - 1)$

$m_2 = 0.125(\frac{29.4}{13.83} - 1)$

$m_2 =0.1407kg$

$m_2 = 140.7 g$

Therefore the mass of the apple was 140.7g

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