Question

Two point particles $A$ and $B$ are placed in line on a frictionless horizontal plane. If particle $A$ (mass 1 kg) is moved with velocity $10m/s$ towards stationary particle $B$ (mass 2 kg) and after collision the two move at an angle of ${45}^o$ with the initial direction of motion, then find
a) velocities of $A$ and $B$ just after collision.
b) coefficient of restitution.

Answer 1

The very first step to solve such problems is to find the line of impact which is along the direction of force applied by $A$ on $B$, resulting the stationary particle of $B$, line of impact can be determined. It this case, the line of impact is along the direction of motion of $B$, i.e. ${45}^o$ with the initial direction of motion of $A$.
a) Let us apply the principle of conservation of momentum.

Along x-direction

$m_A{u}_A=M_A{\nu }_{A}cos{45}^o+m_B{\nu }_{B}cos{45}^o$
or $1\left(10\right)=1\left({\nu }_{A}cos{45}^o\right)+2\left({\nu }_{B}cos{45}^o\right)$
or ${\nu }_A+2{\nu }_B=10\sqrt{2}$

Along y-direction:

$0=m_A{\nu }_{A}sin{45}^o-m_B{\nu }_{B}sin{45}^o$
or $0=1\left({\nu }_{A}sin{45}^o\right)-2\left({\nu }_{B}sin{45}^o\right)$
or ${\nu }_A=2{\nu }_B$
solving the two equations, ${\nu }_A=\frac{10}{\sqrt{2}}m/s$ and ${\nu }_B=\frac{5}{\sqrt{2}}m/s$
b) Coefficient of restitution

$e=\frac{Velocityofseparationalonglineofimpact}{Velocityofapproachalonglineofimpact}$
$=\frac{{\nu }_B-{\nu }_{A}cos{90}^o}{u_{A}cos{45}^o}=\frac{\frac{5}{\sqrt{2}}-0}{\frac{10}{\sqrt{2}}}=\frac{1}{2}$