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3 years ago

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Lonnie pitches a baseball of mass 0.20 kg. The ball arrives at home plate with a speed of 40 m/s and is batted straight back to

Lonnie with a return speed of 60 m/s. If the bat is in contact with the ball for 0.050 s what is the impulse experienced by the ball? A. 360 N.s B. 20 N.s C. 400 N.s D. 9.0 N.s

1 answer:

suter [353]3 years ago

4 0

Answer:

B) 20N.s is the correct answer

Explanation:

The formula for the impulse is given as:

Impulse = change in momentum

Impulse = mass × change in speed

Impulse = m × ΔV

Given:

initial speed = 40m/s

Final speed = -60 m/s (Since the the ball will now move in the opposite direction after hitting the bat, the speed is negative)

mass = 0.20 kg

Thus, we have

Impulse = 0.20 × (40m/s - (-60)m/s)

Impulse = 0.20 × 100 = 20 kg-m/s or 20 N.s

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Answer:

a). $H=2.45m$

b). $H_{max}=1.94m$

Explanation:

For the block that stays on the track, its maximal height is attained when all of the kinetic energy is converted to potential energy

a).

The height for the block on the longer track can by find using this equation:

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Cancel the mass as a factor in each element in the equation

$H=\frac{v_o^2}{2*g}$

$H=\frac{(6.94m/s)^2}{2*9.8m/s^2}$

$H=2.45m$

b).

The other lost some kinetic energy so, use a projectile motion to determine the total height for the other bock:

$E_k=E_p$

$E_k=m*g*H_1$

$E_k=\frac{1}{2}*m*v_o^2-\frac{1}{2}*m*v^2$

$m*g*H_1=\frac{1}{2}*m*(v_o^2-v^2)$

Solve to v'

$v^2=v_o^2-2*g*H_1$

$v=\sqrt{v_o^2-2*g*H_1}=\sqrt{(6.94m/s)^2-2*9.8m/s^2*1.25m}$

$v=4.8m/s$

$H_{max}=H_1+\frac{v^2*sin(50)}{2*g}=1.25m+\frac{(4.8m/s)^2*sin(50)}{2*9.8m/s^2}$

$H_{max}=1.94m$

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Part b)

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Part A)

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