Question

A 20 g ball of clay traveling east at 2.0 m/s collides with a 30 g ball of clay traveling 30° south of west at 1.0 m/s. What are the speed and direction of the resulting 50 g blob of clay?

Short answer

Speed of the resulting clay is 0.41 m/s and direction is $47.18°$

south of west.

Step-by-step solution

Step 1. Given information is :Mass of ball of clay travelling to east = 20gMass of ball of clay travelling to 30° south of west at 1.0 m/s = 30g

We need to find out the speed and direction of the resulting 50 g blob of clay.

Step 2. Expressing the initial velocity of clay 1 and 2 and using law of conservation of momentum.

Let 20g ball of clay be ball 1 and 30g ball of clay be ball 2.

Initial velocities are :

$$\begin{gathered} \overrightarrow{v_1}=v_1\overrightarrow{i}=2m/s \\ \overrightarrow{v_2}=-v_2\cos 30°\overrightarrow{i}-v_2\sin 30°\overrightarrow{j}=(-0.87\overrightarrow{i}-0.5\overrightarrow{j})m/s \\ U\sin glawofconservationofmomentum: \\ Forx-axis, \\ {m}_1{v}_1-m_2{v}_2\cos 30°=(m_1+m_2)v_{f.x} \\ 0.02kg\times 2m/s-0.87m/s\times 0.03kg=(0.02+0.03)kg\times v_{f.x} \\ {v}_{f.x}=0.278m/s \\ Fory-axis, \\ -m_2{v}_2\sin 30°=(m_1+m_2)v_{f.y} \\ -0.5m/s\times 0.03kg=(0.02+0.03)kg\times v_{f.y} \\ {v}_{f.y}=-\frac{0.015}{0.05}m/s \\ {v}_{f.y}=-0.3m/s \\ Resul\tan torthefinalspeedis: \\ {v}_f=\sqrt{{\left(v_{f.x}\right)}^2+{\left(v_{f.y}\right)}^2} \\ {v}_f=\sqrt{{\left(0.278\right)}^2+{\left(-0.3\right)}^2} \\ {v}_f=0.41m/s \\ Anddirection: \\ \tan \theta =\frac{v_{f.y}}{v_{f.x}} \\ \theta =47.18° \end{gathered}$$