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Zigmanuir [339]

1 year ago

7

Aaron Agin nodded off while driving home from play practice this past Sunday evening. His 1500-kg car hit a series of guardrails

while moving at 19.8 m/s. The first guard rail delivered a resistive impulse of 5700 N•s. The second guard rail pushed against his car with a force of 79000 N for 0.12 seconds. The third guard rail collision lowered the car's velocity by 3.2 m/s. Determine the final velocity of the car.

1 answer:

Inessa [10]1 year ago

5 0

Answer: 6.48m/s

Explanation:

First, we know that Impulse = change in momentum

Initial velocity, u = 19.8m/s

Let,

Velocity after first collision = x m/s

Velocity after second collision = y m/s

Also, we know that

Impulse = m(v - u). But then, the question said, the guard rail delivered a "resistive" impulse. Thus, our impulse would be m(u - v).

5700 = 1500(19.8 - x)

5700 = 29700 - 1500x

1500x = 29700 - 5700

1500x = 24000

x = 24000/1500

x = 16m/s

Also, at the second guard rail. impulse = ft, so that

Impulse = 79000 * 0.12

Impulse = 9480

This makes us have

Impulse = m(x - y)

9480 = 1500(16 -y)

9480 = 24000 - 1500y

1500y = 24000 - 9480

1500y = 14520

y = 14520 / 1500

y = 9.68

Then, the velocity decreases by 3.2, so that the final velocity of the car is

9.68 - 3.2 = 6.48m/s

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b) On quadrupling the stiffness constant while other factors are constant:

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<u>Then the new time period:</u>

$T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T$

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