Vectors with a dose of arithmetic :3

AI Thread Summary
The discussion centers on solving for two velocities given their sum, product, and resultant. The equations derived from the problem are P + Q = 15 m/s and P * Q = 56 m²/s², with the resultant R = 13 m/s. The angle between the velocities was successfully calculated to be 60° using the cosine formula. However, there was confusion regarding the final values of P and Q, which were ultimately clarified to be P = 7 m/s and Q = 8 m/s, or vice versa. The conversation highlights the importance of clear notation to avoid misinterpretation.
Lexadis
Messages
11
Reaction score
1

Homework Statement


Two velocities acting at a particular point are such that:
  • The sum of their respective magnitudes is 15m/s
  • The product of their respective magnitudes are 56m2/s2
  • The resultant is 13m/s.
    Find the two velocities and the angle between them.

Homework Equations


R^2 = P^2 + Q^2 + 2PQcos\Theta

The Attempt at a Solution



At first I tried to change the statements into equations:
P + Q = 15m/s
P * Q = 56m^2/s^2
R = 13m/s

And then I used the binomial expressions knowledge to try solving it:
(P+Q)^2 = P^2 + 2PQ+Q^2
15^2 = P^2 + 2*56 + Q^2
225 = P^2 + Q^2 + 112
P^2 + Q^2 = 225 - 112
P^2 + Q^2 = 113

I then substituted the value obtained above for the following equation:
R^2 = P^2 + Q^2 + 2PQcos\Theta
13^2 = 113 + 2 X 56 X cos\Theta
169 - 113 = 112cos\Theta
56 = 112cos\Theta
cos\Theta = 56/112
cos\Theta = 1/2
\Theta = 60°

Through this I could find the value of the angle between them. And also, that P^2 + Q^2 = 113. But I can't seem to find the separate values for P and Q. Any ideas? Thank you :3
 
Last edited:
Physics news on Phys.org
USE THE EQUATION :-
(P-Q)2 = (P+Q)2 - 4PQ
To find the difference between their magnitudes, after then you can simply use elimination method to solve the two equations you got.
 
(P+Q) P= P2 + PQ

15 P = P2 + 56

Solve that quadratic equation.
 
Oh thank you!
I got the final result as P = 7/8, Q = 7/8. Thank you :3
 
Lexadis said:
Oh thank you!
I got the final result as P = 7/8, Q = 7/8. Thank you :3

I think you mean P = (7, 8) m/s and Q = (7, 8) m/s

Writing P and Q the way you did makes it look like a fraction, which is confusing.
 
Lexadis said:
Oh thank you!
I got the final result as P = 7/8, Q = 7/8. Thank you :3

Don't you mean either P=7 and Q=8 or P=8 and Q=7 ?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top