Aproton 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?

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

c because the action was the roller coaster reaches bottom of hill (action) reaction would be blood tends to drain away from head

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