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Afina-wow [57]
3 years ago
7

A rifle bullet with mass 8.00 g and initial horizontal velocity 280 m/s strikes and embeds itself in a block with mass 0.992 kg

that rests on a frictionless surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of 15.0 cm. After the impact, the block moves in SHM. Calculate the period of this motion.
Physics
1 answer:
anyanavicka [17]3 years ago
6 0

Answer:

0.4113772 s

Explanation:

Given the following :

Mass of bullet (m1) = 8g = 0.008kg

Initial horizontal Velocity (u1) = 280m/s

Mass of block (m2) = 0.992kg

Maxumum distance (x) = 15cm = 0.15m

Recall;

Period (T) = 2π√(m/k)

According to the law of conservation of momentum : (inelastic Collison)

m1 * u1 = (m1 + m2) * v

Where v is the final Velocity of the colliding bodies

0.008 * 280 = (0.008 + 0.992) * v

2.24 = 1 * v

v = 2.24m/s

K. E = P. E

K. E = 0.5mv^2

P.E = 0.5kx^2

0.5(0.992 + 0.008)*2.24^2 = 0.5*k*(0.15)^2

0.5*1*5.0176 = 0.5*k*0.0225

2.5088 = 0.01125k

k = 2.5088 / 0.01125

k = 223.00444 N/m

Therefore,

Period (T) = 2π√(m/k)

T = 2π√(0.992+0.008) / 233.0444

T = 2π√0.0042910

T = 2π * 0.0655059

T = 0.4113772 s

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Answer:

Answer is explained in the explanation section.

Explanation:

A)

Solution:

For this to find, we need to calculate the centripetal acceleration on the equator.

The centripetal acceleration of the equator:

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a = 4\pi ^{2}R/T^{2}

By plugging in the values, we get:

a = 4 x (3.14^{2}) x (6.3 x 10^{6}) / 86164.1^{2}

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Hence, this is the centripetal acceleration on the equator. And we also know that, acceleration due to gravity is 9.8 m/s^{2} which is very higher than the centripetal acceleration on the equator.

B) Normal force exerted by chair will always be equal and opposite to the mass times gravitational acceleration (F = mg). Otherwise, I would be thrown away from chair in case the normal force is not equal and opposite or I would be drag down to the earth due to greater mass times gravitational acceleration. Hence, both are equal and opposite.

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9.8 - 0.03 = 9.77

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% diff = 0.306%

Obviously, this is way too low to experience.

D) With the help of same formula as discussed above, we have:

a = 4\pi ^{2}RcosФ/T^{2},

Here, Ф = 44.4°

Just putting the values. we get

a = 0.0241 m/s^{2}

Acceleration while sitting in my chair.

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34kurt

Answer:

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