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3 years ago

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We can compare these two interactions on the basis of impulse (see above), but sometimes, we are more interested in the forces (

human body can withstand very large amount of impulse, if it's delivered over a long time with small forces, but we cannot withstand very large forces lasting over more than a few milliseconds, delivering relatively small impulse). In order to estimate average force from impulse, we need the duration of interaction. Suppose that the contact of the bat with the baseball in (b) lasts for 0.7 milliseconds. What is the magnitude of the average force that the bat exerts on the baseball, for the duration of contact

1 answer:

kiruha [24]3 years ago

7 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$I = 476 \ N \cdot s$

b

$I_1 = 14.21 \ N\cdot s$

c

$F = 20300 \ N$

Explanation:

Considering the first question

From the question we are told that

The force produced is $F = 3400 \ N$

The duration of the punch is $t = 0.14 \ s$

Generally the impulse delivered is mathematically represented as

$I = F * t$

=> $I = 3400 * 0.14$

=> $I = 476 \ N \cdot s$

Considering the second question

The approaching velocity of the ball is $v_b = 45 \ m/s$

The leaving velocity of the ball is $v_l = -53 \ m/s$

The mass of the ball is $m_b = 0.145 \ kg$

Generally the magnitude of the impulse delivered is mathematically represented as

$I_1 = m* v_b - m * v_l$

=> $I_1 = [0.145 * 45] - [0.145 * -53]$

=> $I_1 = 14.21 \ N\cdot s$

Considering the third question

The duration of the impact of the bat is $t _1 = 0.7 \ ms = 0.7 *10^{-3} \ s$

Generally the average force exerted by the bat is mathematically represented as

$F = \frac{I_1}{t_1}$

=> $F = \frac{14.21 }{0.7 *10^{-3}}$

=> $F = 20300 \ N$

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Reason:

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Force exerted on the ball by the bat = $F_{bat}$

Given that the batter hits the ball with the bat, the force exerted by the bat on the ball, $F_{bat}$ , is reacted to by the force of the ball acting on the bat, $F_{ball}$

According to Newton's third law of motion, action and reaction are equal and opposite. Therefore, at the moment when the ball hits the bat, we have;

$\underline{F_{ball}}$ <u> on bat</u> = $\underline{F_{bat}}$ <u> on ball</u>

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$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$

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x = 0 (origin)

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In order to find how long the mortar shell remain in the air, we can use the equation:

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x = 1322 ft is the final position of the shell when it strikes the hill

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Substituting these values into the equation, we find the time of flight of the shell:

$t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s$

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In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

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Instead, the vertical component of the velocity is given by

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$v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s$

Therefore, the final speed of the shell is:

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