Questions
A nuclide at rest emits an -particle. In this process:
a
α- particle moves with large velocity and the nucleus remains at rest
b
both α-particle and nucleus move with equal speed in opposite direction
c
both move in opposite directions but nucleus with greater speed
d
both move in opposite direction but α-particle with greater speed.
detailed solution
Correct option is D
As Fext = 0, hence Ps→ = constantor Pα→+PN→ = constantBecause nuclide was at rest before emission of α particle, hence according to law of conservation of momentum, Pα→+PN→ = 0or mαvα→+ mNvN→ = 0 or vα → = -mNmαvN→Thus, α-particle and the nucleus move in opposite directions. Further, as mN >> mα, hence vα>>vNTalk to our academic expert!
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In the figure given the position-time graph of a particle of mass 0.1 kg is shown. The impulse at t = 2 s is :
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