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jonny [76]
3 years ago
15

A 50-g ball moving at 10 m/s in the +x direction suddenly collides head-on with a stationary ball of mass 100 g. If the collisio

n is perfectly elastic, what is the velocity of each ball immediately after the collision?A. +6.7 m/s, -3.3 m/sB. -6.7 m/s, +3.3 m/sC. +3.3 m/s, -6.7 m/sD. -3.3 m/s, +6.7 m/s
Physics
1 answer:
Savatey [412]3 years ago
6 0

Answer:

<h3>3.33m/s</h3>

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses of the object

u1 and u2 are the velocities

v is the final velocity

Given

m1 = 50g

u1 = 10m/s

m2 = 100g

u2 = 0m/s (stationary ball)

Required

Common velocity v

Substitute

50(10) + 100(0) = (50+100)v

500 = 150v

v = 500/150

v = 3.33m/s

Hence the velocity of each ball immediately after the collision is 3.33m/s

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A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em
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Answer:

a. 0.000002 m

b. 0.00000182 m

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36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

P = \ro g h = 1000 * 10 * 20 = 200000 Pa

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

P_1V_1 = P_2V_2

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

V_1 = Ah

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

P_1Ah_1 = P_2Ah_2

h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m

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