Question

0.500 kg ball traveling at 1.00 m/s to the right hits a second ball of equal mass that is initially at rest. After the perfectly elastic collision, the second ball travels at 0.543 m/s at a 57.2o angle upward from the point of contact.
Draw a before and after diagram of this collision.
Write the momentum equations in both the x and y direction.
Using the information from the previous problem, determine the speed and angle of the first ball.

Answer 1

We write the momentum equations in the x and y direction.

in the x direction

$m_1 \cdot v_1=m_1 \cdot u_1 \cdot cos{(\phi_1)}+m_2 \cdot u_2 \cdot cos{(\phi_2)}$ (1)

in the y direction

$0=-m_1 \cdot u_2 \cdot sin{(\phi_1)}+m_2 \cdot u_2 \cdot sin{(\phi_2)}$ (2)

we write the law of conservation of kinetic energy before and after the collision

$\frac{m_1v_1^2}{2}=\frac{m_1u_1^2}{2}+\frac{m_2u_2^2}{2}$

by condition $m_1=m_2$

Then write

$v_1^2=u_1^2+u_2^2$

where from

$u_1=\sqrt{v_1^2-u_2^2}=\sqrt{1^2-0.543^2}=0.84m/s$

using equation (2), we determine the angle $\phi_1$

$0=-m_1 \cdot u_1 \cdot sin{(\phi_1)}+m_2 \cdot u_2 \cdot sin{(\phi_2)}$

$u_1 \cdot sin{(\phi_1)}=u_2 \cdot sin{(\phi_2)}$

$sin{(\phi_1)}=\frac{u_2 \cdot sin(\phi_2)}{u_1}=\frac{0.543 \cdot sin(57.2^o)}{0.84}=$ 0.543

Then

$\phi_1=\arcsin(0.543)=32.888^o$