0
918views
Two smooth balls collides as shown in figure. If mA= 1 kg, mB=2kg and e= 0.75 Find the velocities after impact.
1 Answer
written 3.9 years ago by |
(Given) mA=1 kg, mB=2kg,Let u and v be the initial and final velocities.uA=3 m/s, uB=1m/s;Let′s take X−axis be the dotted line as the line of impact as shown in fig.
mA.uAx+mB.uBx=mA.vAx+mB.vAx1×3cos30+2×(−1cos45)=1 vAx+2 vBx1.1839=vAx+2 vBx⋯(1)
Coefficient of restitution (e) =vBx−vAxuAx−uBx0.75=vBx−vAx3cos30+1cos450.75=vBx−vAx3.30520.75×3.3052=vBx−vAx2.4789=vBx−vAx⋯(2)Equating (1) and (2), we get,VAx=−1.2580 and VBx=1.2209;
vAy=uAy=3sin30=1.5 and vBy=uBy=−1sin45=−0.7071
After impact, Velocity of ball A =√(VAx)2+(VAy)2=√(−1.2580)2+(1.5)2=1.9577 m/s
αA=tan−1VAyVAx=tan−11.51.2580=50.0146∘
αB=tan−1VByVBx=tan−10.70711.2209=30.0778∘