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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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Two large aluminum plates are separated by a distance of 2.0 cm and are held at a potential difference of 170 V. An electron ent
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To solve this problem it is necessary to apply the concepts related to energy as a function of voltage, load and force, and the definition of Force given by Newton in his second law.

By definition we know that force is equal to

F= ma

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m = mass (at this case of an electron)

a = Acceleration

But we also know that the Energy of an electric object is given by two similar definitions.

1) E= \frac{F}{q}

Where,

F= Force

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2) E = \frac{V}{d}

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Now through the cinematic equations of motion we know that,

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Where,

V_f = Final velocity

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Re-arrange to find v_f,

v_f = \sqrt{v_i^2+2ax}

v_f = \sqrt{2.5*10^5+2*(1.49*10^{15})(0.1*10^{-2})}

v_f = 1.726*10^6 m/s

Therefore the electron's speed when it is 0.1 cm from the negative plate is 1.726*10^6 m/s

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