Chapter 11: Q. 71 (page 291)

FIGURE P11.71 shows a collision between three balls of clay. The three hit simultaneously and stick together. What are the speed and direction of the resulting blob of clay?

Short Answer

Expert verified

Speed of the resulting blob of clay is 0.84 m/s and the direction is $θ = 72.85 °$

Step by step solution

Step 1. Given information is :Mass of three balls of clay are 40g, 30g and 20g with speed and direction as shown in figure.

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

Step 2. Using law of conservation of Momentum

Let the balls of clay with mass 40g, 30g and 20g as 1, 2 and 3 respectively.

$\vec{v_{1}} = v_{1} \cos 45 ° \vec{i} - v_{1} \sin 45 ° \vec{j} = 2.82 m / s \vec{i} - 2.82 m / s \vec{j} \vec{v_{2}} = - v_{2} \vec{i} = - 3 m / s \vec{i} \vec{v_{3}} = v_{3} \vec{j} = 2 m / s \vec{i} - 2.82 m / s \vec{j}$

Using Law of conservation of momentum for both axes,

For x-axis,

$m_{1} v_{1} \cos 45 ° + m_{2} (- v_{2}) = (m_{1} + m_{2} + m_{2}) v_{f i n a l} 0.04 k g \times 2.82 m / s - 0.03 k g \times 3 m / s = 0.09 k g \times v_{f i n a l} v_{f i n a l} = 0.25 m / s$

For y-axis,

localid="1648151136846" $m_{1} (- v_{1} \sin 45 °) + m_{3} v_{3} = (m_{1} + m_{2} + m_{3}) v_{f i n a l} - 0.04 k g \times 2.82 m / s + 0.02 k g \times 2 m / s = 0.09 k g \times v_{f i n a l} v_{f i n a l} = 0.81 m / s S p e e d o f t h e r e s u l t i n g b l o b w i l l b e = \sqrt{{(v_{f i n a l . x})}^{2} + {(v_{f i n a l . y})}^{2}} = \sqrt{{(0.25)}^{2} + {(0.81)}^{2}} = 0.84 m A n d a n g l e i s : \tan θ = \frac{v_{f i n a l . y}}{v_{f i n a l . x}} θ = 72.85 °$