Answer to Question #276502 in Mechanics | Relativity for ivy

Question

An air table has two pucks; one with a mass of 0.50kg and the other with a mass of 1.50kg.  Initially the smaller puck is moving at 0.40m/s, 0° while the larger puck is moving at 0.40m/s, 180°.  After the collision, the smaller puck is moving at 0.79m/s, 196°.  What is the velocity (magnitude and direction) of the larger puck after the collision?

Answer 1

$m_1= 0.5kg$

$v_1= 0.4\frac{m}{s}$

$\alpha_1= 0\degree$

$\vec v_1(v_1\cos \alpha_1,v_1\sin \alpha_1)$

$\vec v_1(0.4;0)$

$m_2= 1.5kg$

$v_2= 0.4\frac{m}{s}$

$\alpha_2= 180\degree$

$\vec v_2(v_2\cos \alpha_2,v_1\sin \alpha_2)$

$\vec v_2(-0.4;0)$

$u_1 = 0.79\frac{m}{s}$

$\alpha'_1= 196\degree$

$\vec u_1(u_1\cos \alpha'_1,u_1\sin \alpha'_1)$

$\vec u_1(-0,76;-0.22)$

$\text{According to the law of momentum:}$

$m_1\vec v_1+ m_2\vec v_2=m_1\vec u_1+m_2\vec u_2$

$0.5*(0.4;0)+1.5*(-0.4;0)= 0.5*(-0.76;-0.22)+1.5*(u_{2x};u_{2y})$

$\vec u_2(-0.013;0.073)$

$u_2=\sqrt{0.013^2+0.073^2}=0.074\frac{m}{s}$

$\cos \alpha'_2=\frac{-0.013}{0.074}=-0.17568$

$\alpha'_2= 100.11\degree$

$\text{Answer: }u_2=0.074\frac{m}{s};\alpha'_2= 100.11\degree$