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A particle moving at speed 0.32 c has momentum P₀. the speed of the particle is increased to 0.72 c then its momentum would be 2.25P₀.
It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.
Mathematically the formula of the momentum is
P = mv
where P is the momentum of the particle
m is the mass of the particle
v is the velocity by which the particle is moving
As given in the question a particle moving at speed 0.32 c has momentum P₀. the speed of the particle is increased to 0.72 c
by using the formula of the momentum and substituting the values of the velocity
P =mv
As mentioned in the question when the particle is moving with 0.32c velocity it has a momentum of P₀
P₀ = m*(0.32c)
If the speed of the particle is increased to 0.72 c the momentum would be
P=m*(0.72c)
by dividing the second equation from the first one
P₀/P = 0.32c/0.72c
P₀/P = 0.44
P =2.25P₀
Thus, the increased momentum of the particle would be 2.25P₀
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