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3 years ago

9

A bat hits a moving baseball. If the bat delivers a net eastward impulse of 1.3 N-s and the ball starts with an initial horizont

al velocity of 3.8 m/s to the west and leaves with a 5.4 m/s velocity to the east, what is the mass of the ball (in grams)?

1 answer:

Georgia [21]3 years ago

3 0

Answer:

$141.30grams$

Explanation:

First, denote our known values;

$J=1.3N-s(+,east)\\u=-3.8ms^-^1(-,west)\\v=5.4ms^-^1(+,east)\\m=?$

Mass is impulse divided by change in velocity:

$m=\frac{J}{v-u}\\=\frac{1.3}{5.4--3.8}\\\\=\frac{1.3}{9.2}\\=0.1413Kgs$

Hence, the mass of the ball is 141.30grams

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A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju

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Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

upward acceleration of the helicopter, $a=5\ m.s^{-2}$

time after the takeoff after which the engine is shut off, $t_a=10\ s$

a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

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where:

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t = time of observation

$h=0+0.5\times 5\times 10^2$

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b)

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

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<u>height fallen freely by Austin:</u>

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

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$t_f=$ time of free fall

$h_f=0+0.5\times 9.8\times 7^2$

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$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

<u>Time taken by the helicopter to fall:</u>

$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

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$250=0+0.5\times 9.8\times t_h^2$

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$t'=t_h-t_f$

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$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

$h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2$

$h'=9.8165\ m$

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$\Delta h=h-(h_f+h')$

$\Delta h=250-(240.1+9.8165)$

$\Delta h=0.0835\ m$

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