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
