Chapter 11: Q 70 (page 291)

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

Expert verified

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 :

$\vec{v_{1}} = v_{1} \vec{i} = 2 m / s \vec{v_{2}} = - v_{2} \cos 30 ° \vec{i} - v_{2} \sin 30 ° \vec{j} = (- 0.87 \vec{i} - 0.5 \vec{j}) m / s U \sin g l a w o f c o n s e r v a t i o n o f m o m e n t u m : F o r x - a x i s, m_{1} v_{1} - m_{2} v_{2} \cos 30 ° = (m_{1} + m_{2}) v_{f . x} 0.02 k g \times 2 m / s - 0.87 m / s \times 0.03 k g = (0.02 + 0.03) k g \times v_{f . x} v_{f . x} = 0.278 m / s F o r y - a x i s, - m_{2} v_{2} \sin 30 ° = (m_{1} + m_{2}) v_{f . y} - 0.5 m / s \times 0.03 k g = (0.02 + 0.03) k g \times v_{f . y} v_{f . y} = - \frac{0.015}{0.05} m / s v_{f . y} = - 0.3 m / s R e s u l \tan t o r t h e f i n a l s p e e d i s : v_{f} = \sqrt{{(v_{f . x})}^{2} + {(v_{f . y})}^{2}} v_{f} = \sqrt{{(0.278)}^{2} + {(- 0.3)}^{2}} v_{f} = 0.41 m / s A n d d i r e c t i o n : \tan θ = \frac{v_{f . y}}{v_{f . x}} θ = 47.18 °$