A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. when the particle lands on the level ground the magnitude of the change in its momentum will be
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mg is the answer of this question
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Answer:
mv
Explanation:
Here the particle follows a projectile motion.
In projectile Motion the velocity along the x-axis does not change.
∴ Change in momentum along x-axis is zero.
we will only consider the velocity along the y-axis to get the change in momentum of the particle.
so,
velocity along the y-axis = vsin45°
|ΔP|= m(Δv)
|ΔP|= m[vsin45°-(-vsin45°)]
|ΔP|=2×mv×sin45°
|ΔP|=2×mv×1/
|ΔP|= mv
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