Grade 12Mechanics

a particle moves in xy plane under the action of force F such that the linear momentum at any time t is P(x)=2cos(t),P(y)=2sin(t) the angle theta b/w P & F @ a t is

guru

8 Years agoGrade 12

2 Answers

Karan Gupta

8 Years ago

P(x) = 2 cos(t), F(x) = dP(x)/dt = -2 sin(t)

P(y) = 2 sin (t), F(y) = dP(y)/dt = 2 cos (t)

Dot product of momentum and Force = F(x)*P(x) + F(y)*P(y) = -4 sin(t)cos(t) + 4 sin(t)cos(t) = 0

Since dot product is zero, angle between the two is 90 degrees

Kushagra Madhukar

5 Years ago

Dear student,

Please find the answer to your question.

P(x) = 2 cos(t), F(x) = dP(x)/dt = -2 sin(t)

P(y) = 2 sin (t), F(y) = dP(y)/dt = 2 cos (t)

Dot product of momentum and Force = F(x)*P(x) + F(y)*P(y) = -4 sin(t)cos(t) + 4 sin(t)cos(t) = 0

Since dot product is zero, angle between the two is 90 degrees

Thanks and regards,

Kushagra

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a particle moves in xy plane under the action of force F such that the linear momentum at any time t is P(x)=2cos(t),P(y)=2sin(t) the angle theta b/w P & F @ a t is

guru

2 Answers

Karan Gupta

Kushagra Madhukar

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