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 , 7 Years ago
Grade 12
2 Answers
Karan Gupta
Last Activity: 7 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
Last Activity: 4 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
Provide a better Answer & Earn Cool Goodies
Enter text here...
LIVE ONLINE CLASSES
Prepraring for the competition made easy just by live online class.
Full Live Access
Study Material
Live Doubts Solving
Daily Class Assignments
Ask a Doubt
Get your questions answered by the expert for free