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The velocity of a body of mass 10 kg increases from 4 m/s to 8 m/s when a force acts on it for 2 s. (a) What is the momentum before the force acts? (b) What is the momentum after the force acts?

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The velocity of a body of mass 10 kg increases from 4 m/s to 8 m/s when a force acts on it for 2 s.

(a) What is the momentum before the force acts?

(b) What is the momentum after the force acts?

(c) What is the gain in momentum per second?

(d) What is the value of the force?

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Mass of the body = 10 kg

Initial velocity u = 4 m/s

Final velocity v = 8 m/s

Time t = 2 s

(a) Momentum before force acts p1 = m x u

= 10 x 4 = 40 kg.m/s

(b) Momentum after force acts p2 = m x v

= 10 x 8 = 80 kg.m/s

(c) Gain in momentum for 2 s = p2 - p1

= 40 kg.m/s

Gain in momentum per second = \(\frac{40}{2}\) = 20 kg.m/s

(d) Acceleration a = \(\frac{v -u}{t}\) = \(\frac{8 -4}{2}\) = 2 m/s2

Force = m x a = 10 x 2 = 20 N

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