13.
An object has a constant momentum of 2.45* 10² kg m/s. Determine the momentum of the object
if its mass decreases to 1/3 of its original value and applied forces causes the speed to increased
to exactly four times of the original. The direction of velocity remains same.
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Given:-
Let initial momentum be P1 = 2.45*10^2 = 245 kg m/s
Let final momentum be P2 = x
Therefore P1 = m1v1 and P2 = m2v2
Final mass = 1/3m1
Final speed = 4v1
Solution:-
P1/P2 = m1v1 / m2v2
245/P2 = m1v1 / (1/3m1)(4v1)
245/P2 = 1 / (1/3)(4)
245/P2 = 1 / (4/3)
245/P2 = 3/4
P2 = (245)4/3
P2 = 326.66
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Let initial momentum be P1 = 2.45*10^2 = 245 kg m/s
Let final momentum be P2 = x
Therefore P1 = m1v1 and P2 = m2v2
Final mass = 1/3m1
Final speed = 4v1
Solution:-
P1/P2 = m1v1 / m2v2
245/P2 = m1v1 / (1/3m1)(4v1)
245/P2 = 1 / (1/3)(4)
245/P2 = 1 / (4/3)
245/P2 = 3/4
P2 = (245)4/3
P2 = 326.66
Hope you like it. Don’t forget to mark me the brainliest.
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