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3 years ago

6. A 0.09 kg arrow hits a target at 22 m/s and penetrates 4 cm before stopping.

1 answer:

8 0

A ) v = v o - a t
0 = 22 - a · t
a · t = 22
d = v o · t - a t²/2
0.04 = 22 t - 22 t / 2
0.04 = 11 t
t = 0.04 : 11 = 0.003636 s
a = 22 / t
a = 6050 m/s²
F = m · a = 0.09 kg · 6050 m/s²
F ( target→arrow) = - 544.5 N
b ) F ( arrow→target ) = 544.5 N
c ) If the speed was doubled: v = 44 m/s;
F = a m
a = 6050 m/s²
a · t = 44
t = 6050 : 0.04
t = 0.007272 s
d = 44 t - 44 t/2 = 22 t
d = 22 · 0.007272
d = 0.16 m = 16 cm

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Answer:

Explanation:

Given

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Using $s=ut+\frac{at^2}{2}$ in x and y direction

$x=v_0_x\times t+\frac{at^2}{2}$

$4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}$

$4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}$

$a=-1.23 m/s^2$

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$y=v_0_y\times t+\frac{a't^2}{2}$

$7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}$

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Final

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K₀ = ½ m v₁²

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$K_{f}$ = ½ (m + M) [m / (m + M) v₁]²

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c) in this case the system is already formed by the two bodies and since there is no rubbing the mechanical energy is conserved, transforming from kinetics to potential

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$Em_{f}$ = U = (m + M) g h

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Em₀ = $Em_{f}$

½ (m + M) v² = (m + M) g h

v = √ 2gh

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h = L-x

h = L - L cos 15

h = L (1- cos 15)

We replace

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