Download Applied Mathematics by Example: Exercises

Solutions
Applied Mathematics by Example: Exercises
26. Q in equilibrium under F1 = 2pi + 4qj, F2 = pi − 10j, F3 = 3i + qj. Hence resultant
F = F1 + F2 + F3 = 0; thus (2p + p + 3)i + (4q − 10 + q)j = 0 – i.e. 3p + 3 = 0
and 5q− 10 = 0 – giving
mag(−2i +
1 = magnitude(F1 ) = √
√ p = −1, q = 2; so F
8j) = ((−2)2 + 82√
) = 68 N; similarly, F2 = ((−1)2 + (−10)2 ) = 101 N and
F3 = (32 + 22 ) = 13 N.
27. Ball A mass m (= 0.15 kg) and velocity v (= 0.2i m s−1 ) has momentum P = mv =
0.03i N s. Ball A strikes ball B; after impact, B has velocity (i + j)/10 and A has
velocity u. Momentum conservation means (m/5)i +√m0 = mu + (m/10)(i + j).
Hence u =
= mag((i +
√ (i − j)/10. A-speed = mag((i − j)/10) = 2/10; B-speed
◦
j)/10) = 2/10. As shown in sketch (below), velocity of A is 45 clockwise from
x-axis; velocity of B is 45◦ anti-clockwise from x-axis, so 90◦ is total angle between
A and B directions of motion.
vB =
1
10 (i
+ j)
vA =
1
10 (i
− j)
Download free ebooks at bookboon.com
120