Answer to Question #99632 in Physics for Kanny

Question

Two students throw playdough in a kindergarten class and collide in a completely inelastic collision. Just before the collision a 2.84 g piece is travelling at 7.80 m/s [N] and a 3.40 g piece is travelling at 6.25 m/s [20°N of E]. What is the velocity of the playdough after the collision?

Answer 1

From the conservation of momentum:

$$(m_1+m_2)v_x=m_2v_2\cos{20}$$

$$(m_1+m_2)v_y=m_1v_1+m_2v_2\sin{20}$$

Thus,

$$(2.84+3.4)v_x=(3.4)(6.25)\cos{20}$$

$$v_x=3.2\frac{m}{s}$$

$$(2.84+3.4)v_y=(2.84)(7.8)+(3.4)(6.25)\sin{20}$$

$$v_y=4.715\frac{m}{s}$$

Magnitude:

$$v=\sqrt{3.2^2+4.715^2}=5.70\frac{m}{s}$$

Direction:

$$\theta=\arctan{\frac{4.715}{3.2}}=55.8\degree N\ of\ E$$