Answer:
(a) linear momentum = 8.75. angular momentum =0. (b) 2.5m/s
Explanation:
A and B are on horizontal and so is the velocity given to A = 3.5m/s i (in i direction which is x direction ), this means that there would be motion only in x direction.
let's first calculate linear momentum.
(A)
p(total)= P_A+P_B = (2.5Kgx3.5m/s)+(1kgx0m/s) = 8.75m/s.
Angular momentum.
L=m*v x r. as the motion is only in x direction and there is no rotation in the system, there fore m*v x r = 0
(B)
Velocity of A and B come from the fact that total linear momentum is conserved.
p_final = P_A+P_B = (mA=mB)v_new = mA*V1+mB*vB.
second term on right is = 0 because B has 0 velocity.
solving for V_new and with the values of all unknown substituted in gives
V_new = (mA*VA)/(mA+mB)= 8.75/3.5= 2.5m/s
The diagram showing the 2 spheres is missing, so i have attached it.
Answer:
A) Linear momentum =(8.75 kg/m.s)i
Angular momentum = (-0.5kg.m²/s)k
B) Va' = (1.5m/s)i and Vb' = (5 m/s)j
Explanation:
First of all, let's find the position of the mass centre;
y' = Σ(mi.yi)/mi = [2.5(0) + 1(0.2)]/(2.5+1) = 0.2/3.5 =0.057143 m
A) Linear momentum is given as;
L = m(a) x v(o) = (2.5 x 3.5)i = (8.75 kg/m.s)i
Angular momentum is given as;
HG = Vector GA x m(a) x v(o)
Where vector GA is the position of the mass centre;
Thus;
HG = 0.057143j x (2.5 x 3.5)i = (-0.5 kg.m²/s) k
B) from conservation of linear momentum;
MaVo = Ma Va' + Mb Vb
2.5 x 3.5 = 2.5Va' + 1 Vb'
8.75 = 2.5Va' + 1 Vb' - - - - eq(1)
Also, for conservation of angular momentum ;
raMaVo = - raMa Va' + rbMb Vb'
from the diagram attached, ra + rb = 0.2
Now, ra is the same as value as that of the centre of the mass.
Thus, ra = 0.057143.
rb = 0.2 - ra = 0.2 - 0.057143 = 0.14286
Thus;
0.057143 x 8.75 = -(0.057143 x 2.5)Va' + (0.14286 x 1) Vb'
0.5 = - 0.14286Va' + 0.14286Vb' - - - eq 2
Solving eq 1 and 2 simultaneously, we get; Va' = 1.5m/s and Vb' = 5 m/s
