Answer to Question #111965 in Mechanics | Relativity for Maya

Question

A ¼kg clay ball rolling E at 10.7m/s collides with a 7/8kg croquet ball on a smooth(almost) friction less surface with this larger ball rolling 10 degrees N of W at 6.3 m/s. Find the resultant velocity and direction and the distance they travel after becoming stuck together if they continue moving for 3.15s. Assume uniform but unknown acceleration.

Answer 1

$$\frac{1}{4}(-10.7,0)+\frac{7}{8}(6.3\cos{10},6.3\sin{10})\\=\left(\frac{1}{4}+\frac{7}{8}\right)(v_x,v_y)$$

$$(v_x,v_y)=(2.45,0.85)\frac{m}{s}$$

$$v=\sqrt{2.45^2+0.85^2}=2.6\frac{m}{s}$$

Direction:

$$\theta=\arctan{\frac{0.85}{2.45}}=19\degree \text{N of W}$$

$$d=0.5vt=0.5(2.6)(3.15)=4.1\ m$$