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bagirrra123 [75]
2 years ago
8

A proton of mass m moving with a speed of 3.0 × 106 m/s undergoes a head on elastic collision with an alpha particle of mass 4m

at rest. What are the velocities of the two particles after the collision?
Physics
1 answer:
Sergio039 [100]2 years ago
4 0

Answer:6.0×10^5m/s

Explanation:

According to the law of conservation of momentum, sum of the momenta of the bodies before collision is equal to the sum of their momenta after collision.

After their collision, the two bodies will move with a common velocity (v)

Momentum = mass × velocity

Let m1 be the mass of the proton = m

Let m2 be the mass of the alpha particle = m2

Let v1 be the velocity of the proton = 3.0×10^6m/s

Let v2 be the velocity of the alpha particle = 0m/s (since the body is at rest).

Using the law,

m1v1 + m2v2 = (m1 + m2)v

m(3.0×10^6) + 4m(0) = (m + 4m)v

m(3.0×10^6) = 5mv

Canceling 'm' at both sides,

3.0×10^6 = 5v

v = 3.0×10^6/5

The common velocity v = 6.0×10^5m/s

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A sample of an unknown substance has a mass of 0.465 kg. if 3,000.0 j of heat is required to heat the substance from 50.0°c to 1
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<h3>What is specific heat capacity?</h3>

The amount of heat required to increase a substance's temperature by one degree Celsius is known as specific heat capacity.

Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.

The given data in the problem is;

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To learn more about the specific heat capacity refer to the link brainly.com/question/2530523

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