Solveeit Logo
Login

Question

Physics Question on Motion in a plane

At what angle must the two forces (x + y) and (x - y) act so that the resultant may be x2+y2\sqrt{x^2 + y^2} ?

A

cos⁡−1[−x2+y22(x2−y2)]\cos^{-1} \Bigg[ - \frac{x^2 + y^2}{2(x^2 - y^2)}\Bigg]

B

cos⁡−1[−2(x2−y2)x2+y2]\cos^{-1} \Bigg[ - \frac{2(x^2 - y^2)}{x^2 + y^2}\Bigg]

C

cos⁡−1[−(x2+y2)(x2−y2)]\cos^{-1} \Bigg[ - \frac{(x^2 + y^2)}{(x^2 - y^2)}\Bigg]

D

cos−1[−(x2−y2)(x2+y2)]cos^{-1} \Bigg[ - \frac{(x^2 - y^2)}{(x^2 + y^2)}\Bigg]

Answer

cos⁡−1[−x2+y22(x2−y2)]\cos^{-1} \Bigg[ - \frac{x^2 + y^2}{2(x^2 - y^2)}\Bigg]

Explanation

Solution

R2=A2+B2+2AB cos θR^2 = A^2 + B^2 + 2AB \, cos \, \theta
Substituting, A=(x+y),B=(x−y) and R=x2+y2,A = (x + y), B = (x - y) \, and \, R = \sqrt{x^2 + y^2},
we get, θ=cos−1[−x2+y22(x2−y2)]\theta = cos^{-1} \Bigg[ - \frac{x^2 + y^2}{2(x^2 - y^2)}\Bigg]