Find angular momentum of particle of mass 0-01 kg- position vector r-10 -6 - meter and moving with a velocity 5 - meter-sec- About the origin-

Momentum p= mv = 0.01× 5î=0.05î Angular momentum = #160; m×( r⃗×v⃗) #160;= 0.01 ×(5î× (10î + 6ĵ)) #160;= 0.01 ×(30k̂) #160;= ...

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Impulsive Forces in Collision

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Question

Find angular momentum of particle of mass $0.01 k g$ , position vector $\to r = (10^i + 6^j)$ meter and moving with a velocity $5^i m e t e r / s e c$ . About the origin.

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0.01 K g

\to r = (10^i + 6^j)

^i m e t r e / s e c

\to r = (3^i + 4^j)

\to v = (5^i - 6^j)

0.05 kg

\to r = (8^i + 4^j) m

2^i m/s

0.05 kg

\to r = (8^i + 4^j) m

2^i m/s

2 k g

t

\to r = 5^i - 2 t^{2}^j .

t = 2 s

k g m^{2} s^{- 1}

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